# Evaluate the following integrals:

Question:

Evaluate the following integrals:

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

Solution:

Given $I=\int \frac{x+2}{\sqrt{x^{2}-1}} d x$

Integral is of form $\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x$

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

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

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

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

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

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

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

Consider $\int \frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}-1}} \mathrm{dx}$

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

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

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

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

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

$=\sqrt{u}=\sqrt{x^{2}-1}$

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

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

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

Then,

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

$=\sqrt{\mathrm{x}^{2}-1}+\cosh ^{-1}(\mathrm{x})+\mathrm{c}$

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