How to evaluate $\int\frac{x}{\sqrt{x^2+x+1}}dx\;?$

Question

How do you integrate, $$\int \frac{x}{\sqrt{x^2+x+1}} dx $$

I am trying to use trig substitution, but I am having trouble finding a perfect square which works.

Answer 1

Complete the square: $$x^2 + x+1 = \left(x+\frac 12\right)^2 + \frac 34 = \left(x+\frac 12\right)^2 + \left(\frac {\sqrt 3}2\right)^2$$

Put $\left(x + \frac 12\right) = \dfrac{\sqrt 3}{2} \tan\theta$.

Answer 2

Hint: $$\frac{x}{x^2+x+1} = \frac{x}{x^2+x+\frac{1}{4} + \frac{3}{4}} = \frac{x}{\left(x+\frac{1}{2}\right)^2 + \frac{3}{4}}$$

Answer 3

$\bf{My\; Solution::}$ Let $$\displaystyle I = \int\frac{x}{\sqrt{x^2+x+1}}dx$$

Let $$\displaystyle x=\left(y-\frac{1}{2}\right)\;,$$ Then $dx = dy$

So $$\displaystyle x^2+x+1 = \left(y^2+\frac{1}{4}-y\right)+\left(y-\frac{1}{2}\right)+ 1= y^2+\frac{3}{4}$$.

So $$\displaystyle I = \int\frac{\left(y-\frac{1}{2}\right)}{\sqrt{y^2+\frac{3}{4}}}dy = \int\frac{y}{\sqrt{y^2+\frac{3}{4}}}dy-\frac{1}{2}\int \frac{1}{\sqrt{y^2+\frac{3}{4}}}dy$$

So $$\displaystyle I = \sqrt{y^2+\frac{3}{4}}-\frac{1}{2}\cdot \ln \left|y+\sqrt{y^2+\frac{3}{4}}\right|+\mathcal{C}\;,$$ Where $\displaystyle y=\left(x-\frac{1}{2}\right)$