Evaluate $\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$

Question

The task is to evaluate $$\int\frac{x^2}{\sqrt{1+x+x^2}}\,dx$$

My best approach has been substitution of $u=x+\frac{1}{2}$, and from there onto (some terrible) trig sub - finally arriving at a messy answer.

Is there a slicker way of doing this?

Answer 1

We have \begin{align} \dfrac{x^2}{\sqrt{1+x+x^2}} & = \dfrac{x^2+x+1}{\sqrt{x^2+x+1}} - \dfrac{x+1}{\sqrt{x^2+x+1}}\\ & = \sqrt{x^2+x+1} - \dfrac12 \dfrac{2x+1}{\sqrt{x^2+x+1}} - \dfrac12 \dfrac1{\sqrt{x^2+x+1}} \end{align}

Answer 2

I give you the answer. With this information, you should have enough information to realize the problem.

$$ \frac{2x-3}{4} \sqrt{1+x+x^2}-\arcsin\left(\frac{1+2x}{\sqrt{3}}\right)/8 $$

Excuse me, I am not used to with this website. I can't give you all the solution; however, this answer give you a good.

Answer 3

As you suggest, complete the square, $$ 1+x+x^2 = (x+1/2)^2 + 3/4 $$ Now, the key is to substitute to make the square root something nice. I'll do a couple of substitutions to make it clearer. Set $y=x+1/2$, so we have $$ \int \frac{(y-1/2)^2}{\sqrt{y^2+3/4}} \, dy $$ Now the denominator looks in the form appropriate for a hyperbolic substitution. In particular, $$ \frac{d}{dx} \arg \sinh{x} = \frac{1}{\sqrt{1+x^2}}, $$ so setting $y=\frac{1}{2}\sqrt{3}\sinh{u}$, you find, ($dy=\frac{1}{2}\sqrt{3}\cosh{u} \, du$) $$ \int \frac{(\frac{1}{2}\sqrt{3}\sinh{u}-\frac{1}{2})^2}{\sqrt{\frac{3}{4} (1+\sinh^2{u})}} \frac{1}{2}\sqrt{3}\cosh{u} \, du = \frac{1}{4}\int (\sqrt{3} \sinh{u}-1)^2 \, du $$ This integral is fairly elementary after using the identity $$ \cosh{2u}=\cosh^2{u}+\sinh^2{u} = 1+2\sinh^2{u} $$ After that, you just have to invert the substitutions, for which you will also need $$ \cosh{u} = \sqrt{1+\sinh^2{u}} $$ and $$ \sinh{2u}=2\sinh{u}\cosh{u}. $$