Exercise 7.4(Integration)

Integrate the functions in Exercises 1 to 23.(Ex 7.4 integration ncert maths solution class 12)

Question 1: $\displaystyle\int\dfrac{3x^2}{x^6+1}dx$

Solution: Let $I=\displaystyle\int\dfrac{3x^2}{x^6+1}dx$

$= \displaystyle\int\dfrac{3x^2}{(x^3)^2+1}dx$

Let $x^3 = t$

$\Rightarrow 3x^2=\frac{dt}{dx}$

$\Rightarrow dx = \frac{dt}{3x^2}dt$

$=\displaystyle\int \dfrac{3x^2}{(t^2+1)}\frac{dt}{3x^2}$

$=\displaystyle\int \dfrac{1}{(t^2+1)}dt$

$=\tan^{-1}t+C$

$I= \tan^{-1}(x^3)+C$

Question 2: $\displaystyle\int\frac{1}{\sqrt{1+4x^2}}dx$

Solution: Let $I = \displaystyle\int\frac{1}{\sqrt{1+4x^2}}dx$

$= \displaystyle\int\frac{1}{\sqrt{1+(2x)^2}}dx$

Let $2x = t$

$\Rightarrow 2 dx = dt$

$\Rightarrow dx = \frac{dt}{2}$

$I = \displaystyle\int\frac{1}{\sqrt{1+(t)^2}}\frac{dt}{2}$

Using formula :

$\displaystyle \int\frac{1}{\sqrt{x^2+a^2}}=\log|x+\sqrt{x^2+a^2}|$

Hence, $I = \frac{1}{2}\log|t+\sqrt{t^2+1}|$

$\Rightarrow I= \frac{1}{2}\log|2x + \sqrt{4x^2+1}|+C$

Question 3: $\displaystyle \int\frac{1}{\sqrt{(2-x)^2+1}}dx$

Solution: Let $I = \displaystyle \int\frac{1}{\sqrt{(2-x)^2+1}}dx$

Let $2-x = t$

$\Rightarrow -dx = dt$

$\Rightarrow dx = -dt$

$I = -\displaystyle \int\frac{1}{\sqrt{(t)^2+1}}dt$

Using formula :

$\displaystyle \int\frac{1}{\sqrt{x^2+a^2}}=\log|x+\sqrt{x^2+a^2}|$

Hence,

$I =-\log|t+\sqrt{t^2+1}|+C$

$= -\log|(2-x)+\sqrt{(2-x)^2+1}|+C$

$= \log\left|\frac{1}{(2-x)+\sqrt{(2-x)^2+1}}\right|+C$

$= \log\left|\frac{1}{(2-x)+\sqrt{4+x^2-4x+1}}\right|+C$

$=\log\left|\frac{1}{(2-x)+\sqrt{x^2-4x+5}}\right|+C$

Question 4: $\displaystyle\int\dfrac{1}{{\sqrt {9 - 25{x^2}} }}dx.$

Solution: Let $I =\displaystyle\int\dfrac{1}{{\sqrt {9 - {(5x)^2}} }}dx.$

Let $5x = t$ ,

$\Rightarrow 5dx = dt$

$\Rightarrow dx = \dfrac{{dt}}{5}$

Now,

$I= \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( t \right)}^2}} }}} \left( {\dfrac{{dt}}{5}} \right)$

$= \dfrac{1}{5}\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( 3 \right)}^2} - {{\left( t \right)}^2}} }}}dt$

$= \dfrac{1}{5}{\sin ^{ - 1}}\left( {\dfrac{t}{3}} \right) + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{a^2} - {x^2}} }}dx} = {{\sin }^{ - 1}}\dfrac{x}{a}} \right]$

$= \dfrac{1}{5}{\sin ^{ - 1}}\left( {\dfrac{{5x}}{3}} \right) + C$

Question 5:- $\displaystyle\int\dfrac{{3x}}{{1 + 2{x^4}}}dx.$

Solution: let $I=\displaystyle\int\dfrac{{3x}}{{1 + {(\sqrt{2}x^2)^2}}}dx.$

Let $\sqrt 2 {x^2} = t$ ,

$2\sqrt 2 xdx = dt$

$xdx = \dfrac{{dt}}{{2\sqrt 2 }}$

Now,

$I= \displaystyle \int {\dfrac{{3x}}{{1 + {{\left( {\sqrt 2 {x^2}} \right)}^2}}}dx}$

$= \displaystyle \int {\dfrac{3}{{1 + {t^2}}}\left( {\dfrac{{dt}}{{2\sqrt 2 }}} \right)}$

$= \dfrac{3}{{2\sqrt 2 }}\displaystyle \int {\dfrac{1}{{1 + {t^2}}}dt}$

$= \dfrac{3}{{2\sqrt 2 }}{\tan ^{ - 1}}t + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{1 + {x^2}}}dx} = {{\tan }^{ - 1}}x + C} \right]$

$I= \dfrac{3}{{2\sqrt 2 }}{\tan ^{ - 1}}\left( {\sqrt 2 {x^2}} \right) + C$

Question 6: $\displaystyle\int\dfrac{{{x^2}}}{{1 - {x^6}}}dx.$

Solution: Let $I =\displaystyle\int\dfrac{{{x^2}}}{{1 - {(x^3)^2}}}dx.$

Let ${x^3} = t,$

$\Rightarrow 3{x^2}dx = dt$

$\Rightarrow {x^2}dx = \dfrac{{dt}}{3}$

Now,

$I = \displaystyle \int {\dfrac{1}{{1 - {t^2}}}\left( {\dfrac{{dt}}{3}} \right)}$

$= \dfrac{1}{3}\displaystyle \int {\dfrac{1}{{1 - {t^2}}}dt}$

$= \dfrac{1}{3}\left( {\dfrac{1}{2}\log |\dfrac{{1 + t}}{{1 - t}}|} \right) + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{{a^2} - {x^2}}}dx} = \dfrac{1}{{2a}}\log |\dfrac{{a + x}}{{a - x}}| + C} \right]$

$=\dfrac{1}{6}\log \left|\dfrac{{1 + {x^3}}}{{1 - {x^3}}}\right| + C$

Question 7: $\displaystyle\int\dfrac{{x - 1}}{{\sqrt {{x^2} - 1} }}dx.$

Solution: Let $I = \displaystyle\int\dfrac{{x - 1}}{{\sqrt {{x^2} - 1} }}dx.$

$= \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} - \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} {\text{ }}$

Let ${x^2} - 1 = t$ ,

Now differentiate both sides,

$\Rightarrow 2xdx = dt$

$\Rightarrow xdx = \dfrac{{dt}}{2}$

$I= \displaystyle \int {\dfrac{1}{{\sqrt t }}\left( {\dfrac{{dt}}{2}} \right)}-\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} {\text{ }}$

$= \dfrac{1}{2}\displaystyle \int {{t^{ - \dfrac{1}{2}}}dt} - \log |x + \sqrt {{x^2} - 1} |+C$

$= \dfrac{1}{2}\left( {\dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}}} \right) - \log |x + \sqrt {{x^2} - 1} |+C$

$= \dfrac{1}{2}\left( {2\sqrt t } \right)- \log |x + \sqrt {{x^2} - 1} |+C$

$= \sqrt t - \log |x + \sqrt {{x^2} - 1} |+C$

$= \sqrt {{x^2} - 1}- \log |x + \sqrt {{x^2} - 1} |+C$

Question 8: $\displaystyle\int\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx.$

Solution: Let $I=\displaystyle\int\dfrac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx.$

$\Rightarrow I= \displaystyle \int {\dfrac{{{x^2}}}{{\sqrt {{{\left( {{x^3}} \right)}^2} + {{\left( {{a^3}} \right)}^2}} }}dx}$

Let ${x^3} = t$ ,

$\Rightarrow 3{x^2}dx = dt$

$\Rightarrow {x^2}dx = \dfrac{{dt}}{3}$

Now,

$= \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( t \right)}^2} + {{\left( {{a^3}} \right)}^2}} }}\left( {\dfrac{{dt}}{3}} \right)}$

$= \dfrac{1}{3}\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( t \right)}^2} + {{\left( {{a^3}} \right)}^2}} }}dt}$

$= \dfrac{1}{3}\log |t + \sqrt {{t^2} + {{\left( {{a^3}} \right)}^2}} | + C{\text{ }}\left[ {\displaystyle \int {\dfrac{{dx}}{{\sqrt {{x^2} + {a^2}} }} = } \log |x + \sqrt {{x^2} + {a^2}} |} \right]$

$I= \dfrac{1}{3}\log |{x^3} + \sqrt {{x^6} + {a^6}} | + C$

Question 9: $\displaystyle\int\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}dx$ .

Solution: Let $I=\displaystyle\int\dfrac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}dx$ .

Let $\tan x = t$ ,

$\Rightarrow {\sec ^2}xdx = dt$

$\Rightarrow dx = \frac{dt}{\sec^2x}$

Now,

$I= \displaystyle \int {\dfrac{{dt}}{{\sqrt {{t^2} + {{\left( 2 \right)}^2}} }}}$

$= \log |t + \sqrt {{t^2} + {{\left( 2 \right)}^2}} | + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}dx} = \log |x + \sqrt {{x^2} + {a^2}} |} \right]$

$= \log |\tan x + \sqrt {{{\tan }^2}x + 4} | + C$

Question 10: $\displaystyle\int\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}dx$ .

Solution : Let $I=\displaystyle\int\dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}dx$

$I= \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( 1 \right)}^2}} }}dx}$

Let x + 1 = t,

Now, differentiate both sides,

dx = dt

Now,

$= \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + {{\left( 1 \right)}^2}} }}dt}$

$= \log |t + \sqrt {{t^2} + 1} | + C{\text{ }}\left[ {\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }}dx} = \log |x + \sqrt {{x^2} + {a^2}} |} \right]$

$= \log |\left( {x + 1} \right) + \sqrt {{{\left( {x + 1} \right)}^2} + 1} | + C$

$I= \log |x + 1 + \sqrt {{x^2} + 2x + 2} | + C$

Question 11: $\displaystyle\int\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx$ .

Solution : Let $I =\displaystyle\int\dfrac{1}{{\sqrt {9{x^2} + 6x + 5} }}dx$

$I= \displaystyle \int {\dfrac{1}{{\sqrt {9{x^2} + 6x + 1 + 4} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {3x + 1} \right)}^2} + {2^2}} }}dx}$

Let $3x + 1 = t,$

$\Rightarrow 3dx = dt$

$\Rightarrow dx = \dfrac{{dt}}{3}$

Now,

$I= \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + {2^2}} }}\left( {\dfrac{{dt}}{3}} \right)}$

$= \dfrac{1}{3}\displaystyle \int {\dfrac{1}{{\sqrt {{t^2} + {2^2}} }}dt}$

$= \dfrac{1}{3}\log |t + \sqrt {{t^2} + {2^2}} | + C$

$= \dfrac{1}{3}\log |\left( {3x + 1} \right) + \sqrt {{{\left( {3x + 1} \right)}^2} + {2^2}} | + C$

$= \dfrac{1}{3}\log |\left( {3x + 1} \right) + \sqrt {9{x^2} + 6x + 5} | + C$

Question 12: $\displaystyle\int\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx$ .

Solution: Let $I = \displaystyle\int\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx$

$\Rightarrow I= \displaystyle \int {\dfrac{1}{{\sqrt {7 - \left( {{x^2} + 6x} \right)} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {7 - \left( {{x^2} + 6x + 9 - 9} \right)} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {7 + 9 - {{\left( {x + 3} \right)}^3}} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {16 - {{\left( {x + 3} \right)}^2}} }}dx}$

$I= \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {{\left( {x + 3} \right)}^2}} }}dx}$

Let, $x + 3 = t$

$\Rightarrow dx = dt$

Now,

$I= \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {{\left( {x + 3} \right)}^2}} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {{4^2} - {t^2}} }}dt}$

$= {\sin ^{ - 1}}\left( {\dfrac{t}{4}} \right) + C$

$\displaystyle \int {\dfrac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} = {\sin ^{ - 1}}\left( {\dfrac{{x + 3}}{4}} \right) + C$

Question 13: $\displaystyle\int\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx$ .

Solution: Let $I=\displaystyle\int\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx$ .

$\Rightarrow I= \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 3x + 2} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 3x + \dfrac{9}{4} - \dfrac{9}{4} + 2} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - \left( {\dfrac{1}{4}} \right)} }}} dx$

$= \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}$

Now, let

$x - \dfrac{3}{2} = t$

$\Rightarrow dx = dt$

Now,

$I= \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( t \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dt}$

$= \log |t + \sqrt {{t^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} | + C$

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx} = \log |\left( {x - \dfrac{3}{2}} \right) + \sqrt {{{\left( {x - \dfrac{3}{2}} \right)}^2} - \dfrac{1}{4}} | + C$

$= \log |\left( {x - \dfrac{3}{2}} \right) + \sqrt {{x^2} - 3x + 2} | + C$

Question 14: $\displaystyle\int\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx$ .

Solution: Let $I =\displaystyle\int\dfrac{1}{{\sqrt {8 + 3x - {x^2}} }}dx$ .

$I= \displaystyle \int {\dfrac{1}{{\sqrt {8 - \left( {{x^2} - 3x} \right)} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {8 - \left( {{x^2} - 3x + \dfrac{9}{4} - \dfrac{9}{4}} \right)} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {8 + \dfrac{9}{4} - \left( {{x^2} - 3x + \dfrac{9}{4}} \right)} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {\left( {\dfrac{{41}}{4}} \right) - {{\left( {{x^2} - \dfrac{3}{2}} \right)}^2}} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {\dfrac{{\sqrt {41} }}{2}} \right)}^2} - {{\left( {x - \dfrac{3}{2}} \right)}^2}} }}dx}$

Let $x - \dfrac{3}{2} = t$

$\Rightarrow dx = dt$

Now,

$I= \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {\dfrac{{\sqrt {41} }}{2}} \right)}^2} - {t^2}} }}dx}$

$= {\sin ^{ - 1}}\dfrac{t}{{\dfrac{{\sqrt {41} }}{2}}} + C$

$= {\sin ^{ - 1}}\dfrac{{2\left( {x - \dfrac{3}{2}} \right)}}{{\sqrt {41} }} + C$

$I= {\sin ^{ - 1}}\dfrac{{2x - 3}}{{\sqrt {41} }} + C$

Question 15: $\displaystyle\int \dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx$ .

Solution: Let $I= \displaystyle\int \dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx$

$\left( {x - a} \right)\left( {x - b} \right) = {x^2} - \left( {a + b} \right)x + ab$

$\left( {x - a} \right)\left( {x - b} \right) = {x^2} - \left( {a + b} \right)x + \dfrac{{{{\left( {a + b} \right)}^2}}}{4} - \dfrac{{{{\left( {a + b} \right)}^2}}}{4} + ab$

$\left( {x - a} \right)\left( {x - b} \right) = {\left[ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right]^2} - \dfrac{{{{\left( {a - b} \right)}^2}}}{4}$

hence,

$\displaystyle \int {\dfrac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt {{{\left[ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right]}^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} }}dx}$

Now, let

$x - \left( {\dfrac{{a + b}}{2}} \right) = t$

$\Rightarrow dx = dt$

Now,

$= \displaystyle \int {\dfrac{1}{{\sqrt {{t^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} }}dx}$

$= \log |t + \sqrt {{t^2} - {{\left( {\dfrac{{a - b}}{2}} \right)}^2}} | + C$

$I= \log |\left\{ {x - \left( {\dfrac{{a + b}}{2}} \right)} \right\} + \sqrt {\left( {x - a} \right)\left( {x - b} \right)} | + C$

Question 16: $\displaystyle\int\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx$ .

Solution: Let $I=\displaystyle\int\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx$

Let $2{x^2} + x - 3 = t,$

$\Rightarrow \left( {4x + 1} \right)dx = dt$

Now,

$\displaystyle \int {\dfrac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx} = \displaystyle \int {\dfrac{1}{{\sqrt t }}} dt$

$= \dfrac{{{t^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}} + C$

$= 2\sqrt t + C$

$= 2\sqrt {2{x^2} + x - 3} + C$

Question 17: $\displaystyle \int\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx$ .

Solution: Let $I=\displaystyle \int\dfrac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx$

$= \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} - 1} }}dx} + \displaystyle \int {\dfrac{2}{{\sqrt {{x^2} - 1} }}dx}$

$\Rightarrow I= \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx} {\text{ }}\left[ {eq.1} \right]$

let ${x^2} - 1 = t$

$\Rightarrow 2xdx = dt$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 1} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{{dt}}{{\sqrt t }}} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 1} }}dx}$

$= \dfrac{1}{2}\left( {2\sqrt t } \right) + 2\log |x + \sqrt {{x^2} - 1} | + C$

$I= \sqrt {{x^2} - 1} + 2\log |x + \sqrt {{x^2} - 1} | + C$

Question 18: $\displaystyle\int\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx$ .

Solution: Let $I=\displaystyle\int\dfrac{{5x - 2}}{{3{x^2} + 2x + 1}}dx$ .

$\Rightarrow I= \displaystyle \int {\dfrac{{5x}}{{3{x^2} + 2x + 1}}dx} - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}$

$\Rightarrow I= \dfrac{5}{6}\displaystyle \int {\dfrac{{6x}}{{3{x^2} + 2x + 1}}dx} - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}$

$\Rightarrow I = \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2 - 2}}{{3{x^2} + 2x + 1}}dx} - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}$

$= \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx - \dfrac{5}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx} } - \displaystyle \int {\dfrac{2}{{3{x^2} + 2x + 1}}dx}$

$\Rightarrow I= \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx} {\text{ }}\left[ {eq.1} \right]$
Let $3{x^2} + 2x + 1 = t$ ,

$\Rightarrow \left( {6x + 2} \right)dx = dt$

$\Rightarrow I= \dfrac{5}{6}\displaystyle \int {\dfrac{{6x + 2}}{{3{x^2} + 2x + 1}}dx} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{x^2} + 2x + 1}}dx}$

$= \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3\left( {{x^2} + \dfrac{2}{3}x} \right) + 1}}dx}$

$= \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3\left( {{x^2} + \dfrac{2}{3}x + \dfrac{1}{9} - \dfrac{1}{9}} \right) + 1}}dx}$

$= \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{{\left( {x + \dfrac{1}{3}} \right)}^2} - \dfrac{1}{3} + 1}}dx}$

$= \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt} - \dfrac{{11}}{3}\displaystyle \int {\dfrac{1}{{3{{\left( {x + \dfrac{1}{3}} \right)}^2} + \dfrac{2}{3}}}dx}$

$= \dfrac{5}{6}\displaystyle \int {\dfrac{1}{t}dt} - \dfrac{{11}}{9}\displaystyle \int {\dfrac{1}{{{{\left( {x + \dfrac{1}{3}} \right)}^2} + {{\left( {\dfrac{{\sqrt 2 }}{3}} \right)}^2}}}dx}$

$= \dfrac{5}{6}\log |t| - \dfrac{{11}}{9}\left( {\dfrac{3}{{\sqrt 2 }}{{\tan }^{ - 1}}\dfrac{{3x + 1}}{{\sqrt 2 }}} \right) + C$

$= \dfrac{5}{6}\log |3{x^2} + 2x + 1| - \dfrac{{11}}{{3\sqrt 2 }}{\tan ^{ - 1}}\left( {\dfrac{{3x + 1}}{{\sqrt 2 }}} \right) + C$

Question 19: $\displaystyle\int\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx$ .

Solution: Let $I=\displaystyle\int\dfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx$

$\Rightarrow I= \displaystyle \int {\dfrac{{6x + 7}}{{\sqrt {{x^2} - 9x + 20} }}}$

$\Rightarrow I= \displaystyle \int {\dfrac{{6x}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}} dx$

$\Rightarrow I= 3\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}$

$= 3\displaystyle \int {\dfrac{{2x - 9 + 9}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}$

$= 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx} + \displaystyle \int {\dfrac{{27}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}$

$= 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx} {\text{ }}\left[ {eq.1} \right]$

Let ${x^2} - 9x + 20 = t$ ,

$\Rightarrow \left( {2x - 9} \right)dx = dt$

$\Rightarrow I= 3\displaystyle \int {\dfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx} + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + 20} }}dx}$

$= 3\displaystyle \int {\dfrac{1}{{\sqrt t }}dt} + 34\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} - 9x + \dfrac{{81}}{4} - \dfrac{{81}}{4} + 20} }}dx}$

$= 3\displaystyle \int {\dfrac{1}{{\sqrt t }}dt} + 34\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x - \dfrac{9}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} }}dx}$

$= 3\left( {2\sqrt t } \right) + 34\log \left|\left( {x - \dfrac{9}{2}} \right) + \sqrt {{{\left( {x - \dfrac{9}{2}} \right)}^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} \right| + C$

$= 6\left( {\sqrt {{x^2} - 9x + 20} } \right) + 34\log \left|\left( {x - \dfrac{9}{2}} \right) + \sqrt {{x^2} - 9x + 20} \right| + C$

Question 20: $\displaystyle\int\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx$ .

Solution: Let $I =\displaystyle\int\dfrac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx$ .

$\Rightarrow I= \displaystyle \int {\dfrac{x}{{\sqrt {4x - {x^2}} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}$

$= - \dfrac{1}{2}\displaystyle \int {\dfrac{{ - 2x}}{{\sqrt {4x - {x^2}} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}$

$= - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x - 4}}{{\sqrt {4x - {x^2}} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}$

$= - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 2} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}$

$= - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx} {\text{ }}\left[ {eq.1} \right]$

Let $4x - {x^2} = t$ ,

$\Rightarrow \left( {4 - 2x} \right)dx = dt$

$I= - \dfrac{1}{2}\displaystyle \int {\dfrac{{4 - 2x}}{{\sqrt {4x - {x^2}} }}dx + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4x - {x^2}} }}dx}$

$= - \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + 4} \displaystyle \int {\dfrac{1}{{\sqrt { - \left( {{x^2} - 4x + 4 - 4} \right)} }}dx}$

$= - \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + 4} \displaystyle \int {\dfrac{1}{{\sqrt {4 - {{\left( {x - 2} \right)}^2}} }}dx}$

$= - \dfrac{1}{2}\left( {2\sqrt t } \right) + 4{\sin ^{ - 1}}\dfrac{{x - 2}}{2} + C$

$= - \sqrt {4x - {x^2}} + 4{\sin ^{ - 1}}\dfrac{{x - 2}}{2} + C$

Question 21: $\displaystyle\int\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx$ .

Solution: Let $I=\displaystyle\int\dfrac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx$

$\Rightarrow I= \displaystyle \int {\dfrac{x}{{\sqrt {{x^2} + 2x + 3} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} + 2x + 3} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2 - 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx - } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx} + 2\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx + } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx} {\text{ [eq}}{\text{.1]}}$

Let ${x^2} + 2x + 3 = t$ ,

$\Rightarrow \left( {2x + 2} \right)dx = dt$

$\Rightarrow dx=\frac{dt}{2x+2}$

$I= \dfrac{1}{2}\displaystyle \int {\dfrac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx + } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 3} }}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + } \displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 2x + 1 + 2} }}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt + } \displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} }}dx}$

$= \dfrac{1}{2}\left( {2\sqrt t } \right) + \log |\left( {x + 1} \right) + \sqrt {{{\left( {x + 1} \right)}^2} + 2} | + C$

$I= \sqrt {{x^2} + 2x + 3} + \log |\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 3} | + C$

Question 22: $\displaystyle\int\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx$ .

Solution: Let $I=\displaystyle\int\dfrac{{x + 3}}{{{x^2} - 2x - 5}}dx$

$= \displaystyle \int {\dfrac{x}{{{x^2} - 2x - 5}}dx} + 3\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{{2x}}{{{x^2} - 2x - 5}}dx} + 3\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{{2x - 2 + 2}}{{{x^2} - 2x - 5}}dx} + 3\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{{2x - 2}}{{{x^2} - 2x - 5}}dx} + 4\displaystyle \int {\dfrac{1}{{{x^2} - 2x - 5}}dx} {\text{ }}\left[ {eq.1} \right]$

Let ${x^2} - 2x - 5 = t$ ,

$\Rightarrow \left( {2x - 2} \right)dx = dt$

$I= \dfrac{1}{2}\displaystyle \int {\dfrac{1}{t}dt} + 4\displaystyle \int {\dfrac{1}{{{x^2} - 2x + 1 - 1 - 5}}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{1}{t}dt} + 4\displaystyle \int {\dfrac{1}{{{{\left( {x - 1} \right)}^2} - {{\left( {\sqrt 6 } \right)}^2}}}dx}$

$= \dfrac{1}{2}\log |t| + 4\left( {\dfrac{1}{{2\sqrt 6 }}} \right)\log \left( {\dfrac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right) + C$

$= \dfrac{1}{2}\log |{x^2} - 2x - 5| + 4\left( {\dfrac{1}{{2\sqrt 6 }}} \right)\log \left( {\dfrac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right) + C$

Question 23: $\displaystyle\int \dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx$ .

Solution: Let $I =\displaystyle\int \dfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx$

$I= \displaystyle \int {\dfrac{{5x}}{{\sqrt {{x^2} + 4x + 10} }}dx} + 3\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}$

$\Rightarrow I= \dfrac{5}{2}\displaystyle \int {\dfrac{{2x}}{{\sqrt {{x^2} + 4x + 10} }}dx} + 3\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}$

$= \dfrac{5}{2}\displaystyle \int {\dfrac{{2x + 4 - 4}}{{\sqrt {{x^2} + 4x + 10} }}dx} + 3\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}$

$= \dfrac{5}{2}\displaystyle \int {\dfrac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx} - 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx} {\text{ }}\left[ {eq.1} \right]$

Let ${x^2} + 4x + 10 = t$ ,

$\Rightarrow \left( {2x + 4} \right)dx = dt$

$\Rightarrow I= \dfrac{5}{2}\displaystyle \int {\dfrac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx} - 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 10} }}dx}$

$= \dfrac{5}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt} - 7\displaystyle \int {\dfrac{1}{{\sqrt {{x^2} + 4x + 4 + 6} }}dx}$

$= \dfrac{5}{2}\displaystyle \int {\dfrac{1}{{\sqrt t }}dt} - 7\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 6 } \right)}^2}} }}dx}$

$= \dfrac{5}{2}\left( {2\sqrt t } \right) - 7\log |\left( {x + 2} \right) + \sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 6 } \right)}^2}} | + C$

$= 5\sqrt {{x^2} + 4x + 10} - 7\log |\left( {x + 2} \right) + \sqrt {{x^2} + 4x + 10} | + C$

Choose the correct answer in Exercises 24 and 25.

Question 24: $\displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 2}}}$ is equals to:

A. $x{\tan ^{ - 1}}\left( {x + 1} \right) + C$

B. ${\tan ^{ - 1}}\left( {x + 1} \right) + C$

C. $\left( {x + 1} \right){\tan ^{ - 1}}\left( x \right) + C$

D. ${\tan ^{ - 1}}\left( x \right) + C$

Solution: the correct answer is B.

Let $I = \displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 2}}}$

$= \displaystyle \int {\dfrac{{dx}}{{{x^2} + 2x + 1 + 1}}}$

$= \displaystyle \int {\dfrac{{dx}}{{{{\left( {x + 1} \right)}^2} + 1}}}$

$= \dfrac{1}{1}{\tan ^{ - 1}}\left( {\dfrac{{x + 1}}{1}} \right) + C$

$= {\tan ^{ - 1}}\left( {x + 1} \right) + C$

Thus, the correct answer is B.

Question 25: $\displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}}$ is equals to:

A. $\dfrac{1}{9}{\sin ^{ - 1}}\left( {\dfrac{{9x - 8}}{8}} \right) + C$

B. $\dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{8x - 9}}{9}} \right) + C$

C. $\dfrac{1}{3}{\sin ^{ - 1}}\left( {\dfrac{{9x - 8}}{8}} \right) + C$

D. $\dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{9x - 8}}{9}} \right) + C$

Solution: the correct answer is B.

Let $I =\displaystyle \int {\dfrac{{dx}}{{\sqrt {9x - 4{x^2}} }}}$

$\Rightarrow I= \displaystyle \int {\dfrac{{dx}}{{\sqrt { - 4\left( {{x^2} - \dfrac{9}{4}x} \right)} }}}$

$= \displaystyle \int {\dfrac{1}{{\sqrt { - 4\left( {{x^2} - \dfrac{9}{4}x + \dfrac{{81}}{{64}} - \dfrac{{81}}{{64}}} \right)} }}dx}$

$= \displaystyle \int {\dfrac{1}{{\sqrt { - 4\left[ {{{\left( {x - \dfrac{9}{8}} \right)}^2} - {{\left( {\dfrac{9}{8}} \right)}^2}} \right]} }}dx}$

$= \dfrac{1}{2}\displaystyle \int {\dfrac{1}{{\sqrt {{{\left( {\dfrac{9}{8}} \right)}^2} - {{\left( {x - \dfrac{9}{8}} \right)}^2}} }}dx}$

$= \dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{x - \dfrac{9}{8}}}{{\dfrac{9}{8}}}} \right) + C$

$= \dfrac{1}{2}{\sin ^{ - 1}}\left( {\dfrac{{8x - 9}}{9}} \right) + C$

Thus, the correct answer is B.

Ex 7.3 integration ncert maths solution class 12

Exercise 7.4(Integration)

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