Mathematics

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

SOLUTION

We have,

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

Put

$t=\sqrt{x+1}$                                        .....(1)

$t^2=x+1$

$x=t^2-1$

D.w.r. to $x$ of equation(1),

$dt=\dfrac{1}{2\sqrt{x+1}}dx$

$dx=2\sqrt{x+1}dt$

Therefore,

$I=2\int{\dfrac{{{t}^{2}+t-1}}{{{t}^{2}}+1}}dt$

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

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

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

$I=2\left[ t+\dfrac{1}{2}\ln \left( {{t}^{2}}+1 \right)-2{{\tan }^{-1}}\left( t \right) \right]+C$

Put the value of $t$, we get

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

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

Hence, the value is $2\left[ \sqrt{x+1}+\dfrac{1}{2}\ln \left( x+2 \right)-2{{\tan }^{-1}}\left( \sqrt{x+1} \right) \right]+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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