# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x$

Solution:

Given $I=\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x$

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

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

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

$\Rightarrow 5 x+3=\lambda(2 x+4)+\mu$

$\therefore \lambda=5 / 2$ and $\mu=-7$

Let $5 x+3=\frac{5}{2}(2 x+4)-7$ and split,

$\Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\int\left(\frac{5(2 x+4)}{2 \sqrt{x^{2}+4 x+10}}-\frac{7}{\sqrt{x^{2}+4 x+10}}\right) d x$

$=5 \int \frac{x+2}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$

Consider $\int \frac{x+2}{\sqrt{x^{2}+4 x+10}} d x$

Let $u=x^{2}+4 x+10 \rightarrow d x=\frac{1}{2 x+4} d u$

$\Rightarrow \int \frac{\mathrm{x}+2}{\sqrt{\mathrm{x}^{2}+4 \mathrm{x}+10}} \mathrm{dx}=\int \frac{1}{2 \sqrt{\mathrm{u}}} \mathrm{du}$

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

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

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

$=\sqrt{u}=\sqrt{x^{2}+4 x+10}$

Consider $\int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$

$\Rightarrow \int \frac{1}{\sqrt{\mathrm{x}^{2}+4 \mathrm{x}+10}} \mathrm{dx}=\int \frac{1}{\sqrt{(\mathrm{x}+2)^{2}+6}} \mathrm{dx}$

Let $u=\frac{x+2}{\sqrt{6}} \rightarrow d x=\sqrt{6} d u$

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

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

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

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

$=\sinh ^{-1}\left(\frac{x+2}{\sqrt{6}}\right)$

Then,

$\Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=5 \int \frac{x+2}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x$

$=5 \sqrt{x^{2}+4 x+10}-7 \sinh ^{-1}\left(\frac{x+2}{\sqrt{6}}\right)+c$

$\therefore \mathrm{I}=\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} \mathrm{dx}=5 \sqrt{x^{2}+4 x+10}-7 \sinh ^{-1}\left(\frac{x+2}{\sqrt{6}}\right)+c$