Evaluate the following integrals:
$\int \frac{x}{\sqrt{x^{2}+x+1}} d x$
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,
$\Rightarrow \int \frac{x}{\sqrt{x^{2}+x+1}} d x=\frac{1}{2} \int \frac{2 x+1}{\sqrt{x^{2}+x+1}} d x-\frac{1}{2} \int \frac{1}{\sqrt{x^{2}+x+1}} d x$
$=\sqrt{x^{2}+x+1}-\frac{\sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)}{2}+c$
$\therefore I=\int \frac{x}{\sqrt{x^{2}+x+1}} d x=\sqrt{x^{2}+x+1}-\frac{\sinh ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)}{2}+c$
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