Evaluate the following integrals:

Solution:

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

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

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

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

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

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

Let $2 x+3=2 x+4-1$ and split,

$\Rightarrow \int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x=\int\left(\frac{2 x+4}{\sqrt{x^{2}+4 x+5}}-\frac{1}{\sqrt{x^{2}+4 x+5}}\right) d x$

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

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

Let $\mathrm{u}=\mathrm{x}^{2}+4 \mathrm{x}+5 \rightarrow \mathrm{dx}=\frac{1}{2 \mathrm{x}+4} \mathrm{du}$

$\Rightarrow \int \frac{x+2}{\sqrt{x^{2}+4 x+5}} d x=\int \frac{1}{2 \sqrt{u}} d u$

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

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

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

$=\sqrt{u}=\sqrt{x^{2}+4 x+5}$

Consider $\int \frac{1}{\sqrt{x^{2}+4 x+5}} d x$

$\Rightarrow \int \frac{1}{\sqrt{x^{2}+4 x+5}} d x=\int \frac{1}{\sqrt{(x+2)^{2}+1}} d x$

Let $u=x+2 \rightarrow d x=d u$

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

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

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

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

Then,

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

$=2 \sqrt{x^{2}+4 x+5}-\sinh ^{-1}(x+2)+c$

$\therefore I=\int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x=2 \sqrt{x^{2}+4 x+5}-\sinh ^{-1}(x+2)+c$