# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{x+2}{\sqrt{x^{2}+2 x+3}}$

Solution:

Given $I=\int \frac{x+2}{\sqrt{x^{2}+2 x+3}} d x$

Integral is of form $\int \frac{\mathrm{px}+\mathrm{q}}{\sqrt{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}}} \mathrm{dx}$

Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$

$\Rightarrow \mathrm{px}+\mathrm{q}=\lambda(2 \mathrm{ax}+\mathrm{b})+\mu$

$\Rightarrow \mathrm{x}+2=\lambda(2 \mathrm{x}+2)+\mu$

$\therefore \lambda=1 / 2$ and $\mu=1$

Let $x+2=1 / 2(2 x+2)+1$ and split,

$\Rightarrow \int \frac{\mathrm{x}+2}{\sqrt{\mathrm{x}^{2}+2 \mathrm{x}+3}} \mathrm{dx}=\int\left(\frac{2 \mathrm{x}+2}{2 \sqrt{\mathrm{x}^{2}+2 \mathrm{x}+3}}+\frac{1}{\sqrt{\mathrm{x}^{2}+2 \mathrm{x}+3}}\right) \mathrm{dx}$

$=\int \frac{\mathrm{x}+1}{\sqrt{\mathrm{x}^{2}+2 \mathrm{x}+3}} \mathrm{dx}+\int \frac{1}{\sqrt{\mathrm{x}^{2}+2 \mathrm{x}+3}} \mathrm{dx}$

Consider $\int \frac{x+1}{\sqrt{x^{2}+2 x+3}} d x$

Let $u=x^{2}+2 x+3 \rightarrow d x=\frac{1}{2 x+2} d u$

$\Rightarrow \int \frac{\mathrm{x}+1}{\sqrt{\mathrm{x}^{2}+2 \mathrm{x}+3}} \mathrm{dx}=\int \frac{1}{2 \sqrt{\mathrm{u}}} \mathrm{du}$

$=\frac{1}{2} \int \frac{1}{\sqrt{\mathrm{u}}} \mathrm{du}$

We know that $\int \mathrm{x}^{\mathrm{n}} \mathrm{dx}=\frac{\mathrm{x}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}$

$\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{u}} d u=\frac{1}{2}(2 \sqrt{u})$

$=\sqrt{u}=\sqrt{x^{2}+2 x+3}$

Consider $\int \frac{1}{\sqrt{x^{2}+2 x+3}} d x$

$\Rightarrow \int \frac{1}{\sqrt{x^{2}+2 x \mp 3}} d x=\int \frac{1}{\sqrt{(x+1)^{2}+2}} d x$

Let $u=\frac{x+1}{\sqrt{2}} \rightarrow d x=\sqrt{2} d u$

$\Rightarrow \int \frac{1}{\sqrt{(x+1)^{2}+2}} d x=\int \frac{\sqrt{2}}{\sqrt{2 u^{2}+2}} d u$

$=\int \frac{1}{\sqrt{u^{2}+1}} d u$

We know that $\int \frac{1}{\sqrt{x^{2}+1}} d x=\sinh ^{-1} x+c$

$\Rightarrow \int \frac{1}{\sqrt{u^{2}+1}} d u=\sinh ^{-1}(u)$

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

Then,

$\Rightarrow \int \frac{x+2}{\sqrt{x^{2}+2 x+3}} d x=\int \frac{x+1}{\sqrt{x^{2}+2 x+3}} d x+\int \frac{1}{\sqrt{x^{2}+2 x+3}} d x$

$=\sqrt{x^{2}+2 x+3}+\sinh ^{-1}\left(\frac{x+1}{\sqrt{2}}\right)+c$

$\therefore \mathrm{I}=\int \frac{\mathrm{x}+2}{\sqrt{\mathrm{x}^{2}+2 \mathrm{x}+3}} \mathrm{dx}=\sqrt{\mathrm{x}^{2}+2 \mathrm{x}+3}+\sinh ^{-1}\left(\frac{\mathrm{x}+1}{\sqrt{2}}\right)+\mathrm{c}$