How can this indefinite integral be solved without partial fractions? 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} 
 A: 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 $-\ln\sqrt{\frac{x^2}{x^2+1}}-\frac2{x^2}+C$
A: I don't think it can be evaluated easily without partial fractions (see Mike's answer, though). After your edit, I see you set things up correctly with the integral you're using partial fractions on. But you only have three equations (which I haven't checked).  You have four unknowns. Find another equation. 
As for the first integral, though, it should be $3\int {dx\over x^3+x}$...
Let's try a decomposition at the start without splitting the integral up first (it would be just as easy):
$$
{3x^2+4\over x^3(x^2+1)}= {A\over x}+{B\over x^2}+{C\over x^3}+{Dx+E\over x^2+1}
$$
Which gives
$$
3x^2+4=Ax^2(x^2+1)+Bx(x^2+1)+C(x^2+1)+(Dx+E)x^3.
$$
Let's figure out what we can be giving $x$ particular values:
Setting $x=0$ gives $C=4$. 
No more nice values.  But we can write now
$$
3x^2+4=Ax^2(x^2+1)+Bx(x^2+1)+4(x^2+1)+(Dx+E)x^3.
$$
Expanding the right hand side of the above and putting it in standard form:
$$
3x^2 +4=(A+D)x^4+(E+B)x^3+(A+4)x^2+Bx+4
$$
So, equating coefficients:
$$\eqalign{
A&=-D\cr
E&=-B\cr
A&=-1\cr
B&=0;\cr
}
$$
whence $A=-1$, $D=1$, $E=B=0$, and (from before) $C=4$.
So,
$$
{3x^2+4\over x^3(x^2+1)}= {-1\over x}+ {4\over x^3}+{ x \over x^2+1}.
$$
A: The demoninator factors into $x^3(x^2+1)$, so to use partial fractions you will end up with $\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{Dx+E}{x^2+1}$.
