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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