At first I had no doubt that I will have to use partial fractions on this integral:
\begin{equation}
\int \frac{3x^2+4}{x^5+x^3}dx
\end{equation}
I split it into two integrals and one of them give me back this equation:
\begin{equation}
\ A(x^2+1)+Bx(x^2+1)+Cx^2(x^2+1)+Dx^4+Ex^3 =1
\end{equation}
Then if I check what happens if $x = 0$ I find out that $A = 1$ after that I checked what happens then $ x =1 \ \ x=-1 \ \ x=2 $ my results there :
\begin{equation}
\ 2B+E=0 \\
\ 2C+D=-1 \\
\ E+2D=2
\end{equation}
I am realizing that I can joggle variables in any fashion I like I will not be get their values.
So using partial fractions is not effective for this particular case. What other way could you suggest of handling this equation?
P.S. I may have made a calculation mistake. In that case I am sorry for wasting your time but I would appreciate if you could point out my mistake.
EDIT: I wrote 'I split it into two integrals' it seams it needs to be shown:
\begin{equation}
3\int \frac{dx}{x^2+1}dx +4 \int \frac{dx}{x^5+x^3}dx = \\
\ \arctan x +4(\int \frac{A}{x^3}dx+\int \frac{B}{x^2}dx+\int \frac{C}{x}dx+\int \frac{Dx+E}{x^2+1}dx)
\end{equation}
Best Answer
To answer the question as asked, it appears given the factor of $1+x^2$ in the denominator, if you want to avoid partial fractions, the substituion
$x=\tan u,dx=\sec^2udu$
looks like it would simplify the denominator.
$\int\dfrac{3x^2+4}{x^5+x^3}dx=\int\dfrac{3\tan^2u+4}{\tan^3u(\tan^2u+1)}\sec^2udu=$
$\int\dfrac{3\tan^2u+4}{\tan^3u\sec^2u}\sec^2udu=$
$\int\dfrac{3\tan^2u+4}{\tan^3u}du=$
$\int3\cot u+4\cot^3udu=\int3\cot u+4\cot u(\csc^2u-1)du=$
$\int-\cot u+4\cot u\csc^2udu=$
$-\ln|\sin u|-2\cot^2u+C$
That first resubstitution is going to be ugly.
$1+\frac1{x^2}=\csc^2u,\dfrac1{1+\frac1{x^2}}=\frac{x^2}{x^2+1}=\sin^2u$
So it looks like we have our result as $-\ln\sqrt{\frac{x^2}{x^2+1}}-\frac2{x^2}+C$