# Evaluate the following integrals:

Question:

Evaluate the following integrals:

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

Solution:

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

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

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

$\Rightarrow p x+q=\lambda(2 a x+b)+\mu$

$\Rightarrow 2 x+1=\lambda(2 x+2)+\mu$

$\therefore \lambda=1$ and $\mu=-1$

Let $2 x+1=2 x+2-1$ and split,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Then,

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

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

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