Evaluate the following integrals:

Question:

Evaluate the following integrals:

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

Solution:

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

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

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

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

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

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

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

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

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

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

Let $\mathrm{u}=\mathrm{x}^{2}+1 \rightarrow \mathrm{dx}=\frac{1}{2 \mathrm{x}} \mathrm{du}$

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

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

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

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

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

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

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

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

Then,

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

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

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

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