# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{x+\sqrt{x+1}}{x+2} d x$

Solution:

The given equation can be written as

$\Rightarrow \int \frac{x}{x+2} d x+\int \frac{\sqrt{x+1}}{x+2} d x$

First integration be $l 1$ and second be $l 2$.

$\Rightarrow$ For l1

Add and subtract 2 from the numerator

$\Rightarrow \int \frac{x+2-2}{x+2}$

$\Rightarrow \int \frac{x+2}{x+2} \cdot d x-\int \frac{2}{x+2} \cdot d x$

$\Rightarrow \int d x-2 \int \frac{d x}{x+2}$

$\Rightarrow x-2 \ln |x+2|+c 1$

$\therefore l1=x-2 \ln |x+2|+c 1$

For 12

$\Rightarrow \int \frac{\sqrt{x+1}}{x+2} d x$

Assume $x+1=t$

$d t=d x$

$\Rightarrow \int \frac{\sqrt{t}}{t+1} d t$

Substitute $u=\sqrt{t}$

$\mathrm{dt}=2 \sqrt{\mathrm{t}} \cdot \mathrm{d} \mathrm{u}$

$t=u^{2}$

$\Rightarrow 2 \int \frac{\mathrm{u}^{2}}{\mathrm{u}^{2}+1} \mathrm{du}$

Add and subtract 1 in the above equation:

$\Rightarrow 2 \int \frac{u^{2}+1-1}{u^{2}+1} d u$

$\Rightarrow 2 \int \frac{u^{2}+1}{u^{2}+1} d u-\int \frac{1}{u^{2}+1} d u$

$\Rightarrow 2 \int d u-\int \frac{1}{u^{2}+1} d u$

$\Rightarrow 2 u-\tan ^{-1}(u)+c 2$

But $u=\sqrt{t}$

$\therefore 2 \sqrt{t}-\tan ^{-1}(\sqrt{t})+c 2$

Also $t=x+1$

$\therefore 2 \sqrt{(} x+1)-\tan ^{-1}(x+1)+c 2$

$l=11+12$

$\therefore 1=x-2 \ln |x+2|+c 1+2 \sqrt{(} x+1)-\tan ^{-1}(x+1)+c 2$

$l=x-2 \ln |x+2|+2 \sqrt{(} x+1)-\tan ^{-1}(x+1)+c$