Integrating a Rational Function

Question

I am studying for a test and I am trying to evauate the integral below. I know how to simplify it with partial fractions, but when I try to solve it, I cannot seem to find a substitution that will simplify it enough to solve in reasonably quick . I plugged it into wolfram and as usual it doesn't give a quick way either. If anybody has a way this integral can be solved quickly, as there will be a time crunch on my test. $$\int\frac{x^3+x+2}{x^4+2x^2+1}dx$$

Answer 1

From partial fractions and two variable substitutions $$\begin{align} \int \frac{x^3+x+2}{x^4+2x^2+1}dx &= \int\frac{x}{x^2+1}dx + 2\int\frac{1}{x^2+1}\frac{1}{x^2+1}dx \\ &= \frac{1}{2}\int\frac{du}{u}+\int\frac{1}{(\tan v)^2+1}dv \end{align}$$

where $u=x^2+1$ and $v=\arctan x$ (recall that $\frac{d}{dx}\arctan x = \frac{1}{1+x^2} $). Now since $\tan^2v+1 = \sec^2 v = 1/\cos^2 v$,

$$\begin{align} \int \frac{x^3+x+2}{x^4+2x^2+1}dx &= \frac{\ln u}{2} + 2\int \cos^2(v)\, dv \\ &= \frac{\ln u}{2}+2\left(\frac{\sin v \cos v}{2} +\frac{v}{2}\right) \\ &= \frac{\ln \left(x^2+1\right)}{2}+\sin \left(\arctan x\right) \cos \left(\arctan x\right) +\arctan x \\ &= \frac{1}{2}\ln \left(x^2+1\right) + \frac{x}{x^2+1} + \arctan x \end{align}$$ Hope this helps.

Answer 2

Try rewriting your integral like this:

$\int \frac{x(x^2+1)+2}{(x^2+1)^2} dx$