Evaluating $\int [(5x^3-3x^2+7x-3)/(x^2+1)^2] dx$ Can I get hints on how to factor? 
Should I break down by term, i.e. $$\frac{5x^3}{(x^2+1)^2}\text{...?}$$ 
 A: Since the degree of the numerator is less than the degree of the denominator, you can immediately apply the method of partial fractions to decompose the integrand: You're looking for numbers $A, B, C, D$ such that,
$$\frac{5x^3 - 3x^2 + 7x - 3}{(x^2 + 1)^2} = \frac{Ax + B}{(x^2 + 1)^2} + \frac{Cx + D}{x^2 + 1}$$
which you can find, for example, by first clearing denominators.
Then, you can deal with each of the pieces separately, each of which is a standard exercise: To handle the terms
$$\frac{Ax}{(x^2 + 1)^2} \qquad \text{and} \qquad \frac{Cx}{x^2 + 1},$$
use a $u$-substitution, and to handle the other two, use a trigonometric substitution.

 The partial fractions decomposition of the rational expression here turns out to be $$\frac{2x}{(x^2 + 1)^2} + \frac{5x-3}{x^2 + 1},$$ so $B = 0$. In particular this means you don't have to evaluate $$\int \frac{dx}{(x^2 + 1)^2},$$ which is easily the most involved of the four integrals.

A: Partial fractions are the reasonable thing to do. We could trot out the machinery, but it can be done at a glance.  
Imagine dividing. The $5x^3+7x$ part gives us $\frac{5x}{x^2+1}+\frac{2x}{(x^2+1)^2}$. The rest gives us $-\frac{3}{x^2+1}$.
The integral can now be just written down. If one wants to show detail, for the first two terms we use the substitution $u=x^2+1$.
