Evaluate the following integrals: x+ x+1 x+2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}+\sqrt{\text{x}+1}}{\text{x}+2}\text{ dx}$

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Answer

We have,
$\text{I}=\int\frac{\text{x}+\sqrt{\text{x}+1}}{\text{x}+2}\text{ dx}$
Let $\text{x}+1=\text{t}^2$
Differentiating both sides we get
$\text{dx}=2\text{t dt}$
Now, integration becomes
$\text{I}=\int\frac{(\text{t}^2-1+\text{t})}{\text{t}^2+1}2\text{t dt}$
$=2\int\frac{\text{t}^3+\text{t}^2-\text{t}}{\text{t}^2+1}\text{ dt}$
$=2\int\frac{\text{t}^3+\text{t}-\text{t}+\text{t}^2+1-1-\text{t}}{\text{t}^2+1}\text{ dt}$
$=\int\frac{\text{t}^3+\text{t}+\text{t}^2+1-\text{t}-\text{t}-1}{\text{t}^2+1}\text{ dt}$
$=2\int\frac{\text{t}^3+\text{t}}{\text{t}^2+1}+2\int\frac{\text{t}^2+1}{\text{t}^2+1}+2\int\frac{-2\text{t}-1}{\text{t}^2+1}\text{ dt}$
$=2\int\text{t dt}+2\int\text{dt}-2\int\frac{2\text{t}}{\text{t}^2+1}\text{ dt}-2\int\frac{1}{\text{t}^2+1}\text{ dt}$
$=\text{t}^2+2\text{t}-2\log\big|\text{t}^2+1\big|-2\tan^{-1}\text{t}+\text{C}$
$=(\text{x}+1)+2\sqrt{\text{x}+1}-2\log|\text{x}+2|-2\tan^{-1}\sqrt{\text{x}+1}+\text{C}$

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