Evaluate the following integrals:

Solution:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Then,

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

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

$\therefore I=\int \frac{x-1}{\sqrt{x^{2}+1}} d x=\sqrt{x^{2}+1}-\sinh ^{-1}(x)+c$