# Evaluate the following integrals:

Question:

Evaluate the following integrals:

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

Solution:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Let $\mathrm{u}=\frac{2 \mathrm{x}+1}{\sqrt{3}} \rightarrow \mathrm{dx}=\frac{\sqrt{3}}{2} \mathrm{du}$

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

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

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

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

$=\sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)$

Then,

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

$=\sqrt{x^{2}+x+1}-\frac{\sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)}{2}+c$

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