# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{x}{\sqrt{x^{2}+6 x+10}} d x$

Solution:

Given $I=\int \frac{x}{\sqrt{x^{2}+6 x+10}} d x$

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

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

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

$\Rightarrow \mathrm{x}=\lambda(2 \mathrm{x}+6)+\mu$

$\therefore \lambda=1 / 2$ and $\mu=-3$

Let $x=1 / 2(2 x+6)-3$ and split,

$\Rightarrow \int \frac{x}{\sqrt{x^{2}+6 x+10}} d x=\int\left(\frac{2 x+6}{2 \sqrt{x^{2}+6 x+10}}-\frac{3}{\sqrt{x^{2}+6 x+10}}\right) d x$

$=\int \frac{x+3}{\sqrt{x^{2}+6 x+10}} d x-3 \int \frac{1}{\sqrt{x^{2}+6 x+10}} d x$

Consider $\int \frac{x+3}{\sqrt{x^{2}+6 x+10}} d x$

Let $u=x^{2}+6 x+10 \rightarrow d x=\frac{1}{2 x+6} d u$

$\Rightarrow \int \frac{\mathrm{x}+3}{\sqrt{\mathrm{x}^{2}+6 \mathrm{x}+10}} \mathrm{dx}=\int \frac{1}{2 \sqrt{\mathrm{u}}} \mathrm{du}$

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

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

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

$=\sqrt{\mathrm{u}}=\sqrt{\mathrm{x}^{2}+6 \mathrm{x}+10}$

Consider $\int \frac{1}{\sqrt{x^{2}+6 x+10}} \mathrm{dx}$

$\Rightarrow \int \frac{1}{\sqrt{x^{2}+6 x+10}} d x=\int \frac{1}{\sqrt{(x+3)^{2}+1}} d x$

Let $u=x+3 \rightarrow d x=d u$

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

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

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

$=\sinh ^{-1}(x+3)$

Then,

$\Rightarrow \int \frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+6 \mathrm{x}+10}} \mathrm{dx}=\int \frac{\mathrm{x}+3}{\sqrt{\mathrm{x}^{2}+6 \mathrm{x}+10}} \mathrm{dx}-3 \int \frac{1}{\sqrt{\mathrm{x}^{2}+6 \mathrm{x}+10}} \mathrm{dx}$

$=\sqrt{\mathrm{x}^{2}+6 \mathrm{x}+10}-3 \sinh ^{-1}(\mathrm{x}+3)+\mathrm{c}$

$\therefore \mathrm{I}=\int \frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+6 \mathrm{x}+10}} \mathrm{dx}=\sqrt{\mathrm{x}^{2}+6 \mathrm{x}+10}-3 \sinh ^{-1}(\mathrm{x}+3)+\mathrm{c}$