How do I know when to use partial fractions or long divison with this integral? $ \int {{x^4+1} \over {x(x^2+1)^2}} dx$ $$ \int {{x^4+1} \over {x(x^2+1)^2}} dx$$
Is there a method to determine which way is better?
 A: The degree of the denominator is greater than the degree of the numerator, so there is no need for preliminary division before using partial fractions.
However, here, instead of trotting out the usual partial fractions machinery,  I would note that $x^4+1=(x^2+1)^2-2x^2$, so we want to integrate 
$$\frac{1}{x}-\frac{2x}{(x^2+1)^2},$$
and this is straightforward. 
A: Hint: start with the substitution $u = x^2$:
$$
\int \frac{x^4+1}{x(x^2+1)^2}dx = \frac 12  \int \frac{u^2+1}{u(1+u)^2} du
$$
A: $$\int {{x^4+1} \over {x \cdot (x^2+1)^2}} \ dx$$
Lets manipulate the integrand into a better form...
$${{x^4+1} \over {x \cdot (x^2+1)^2}}$$
Note that the numerator can be simplified to 
$${{(x^2+1)^2+C} \over {x \cdot (x^2+1)^2}}$$
Now we solve for C, and get
$$C=-2x^2$$
$${{(x^2+1)^2-2x^2} \over {x \cdot (x^2+1)^2}}={{1} \over {x}}-{{2 \cdot x} \over {(x^2+1)^2}}$$
Using this information, the integral becomes...
$$\int {{1} \over {x}}-{{2 \cdot x} \over {(x^2+1)^2}} \ dx$$
The left part is almost trivial to evaluate, for the right hand side, use $u=x^2$
We end up with...
$$\int {{x^4+1} \over {x \cdot (x^2+1)^2}} \ dx=\ln(x)+{1 \over {x^2+1}}+C$$
Where C is now the arbitrary constant...
To answer your actual question. Determining whether or not to use partial fractions is easy. If u-substitution can't be used on the rational function, you can use partial fractions, but you have to make sure the numerator is of a smaller degree than the denominator. However, there will always be methods that work better than partial fractions, but few will be as general...
