Evaluate $\int_{1}^{e}\frac{u}{u^3+2u^2-1}du.$ I'm trying to solve
$$\int_{1}^{e}\frac{u}{u^3+2u^2-1}du.$$
My first approach was to factorise and then do a partial integration. However the factorisation $(u+1)\left(u+\frac{1}{2}-\frac{\sqrt{5}}{2} \right)\left(u+\frac{1}{2}+\frac{\sqrt{5}}{2} \right)$ leads me to heavy calculations. How would you proceed to solve this?
This is a continued calculation of Solving $\int_0^1 \frac{dx}{e^x-e^{-2x}+2}$ with substitution
 A: As $$u^3+2u^2-1=(u+1)(u^2+u-1)$$
Write,
$$\frac u{u^3+2u^2-1}=\frac A{u+1}+B\frac{\dfrac{d(u^2+u-1)}{du}}{u^2+u-1}+\frac C{u^2+u-1}$$
$$u=A(u^2+u-1)+(2Bu+B)(u+1)+C(u+1)$$
Set $\displaystyle u+1=0$ to find $A$
Now comparing the coefficients of $u^2$, $0=A+2B\iff B=?$
Again comparing the constants $0=-A+B+C\implies C=?$
Observe that the second integral $\displaystyle\int B\frac{\dfrac{d(u^2+u-1)}{du}}{u^2+u-1}du=B\int\frac{d(u^2+u-1)}{u^2+u-1}=?$
For the third,  $\displaystyle\int\frac C{u^2+u-1}=4C\int\frac{du}{(2u+1)^2-(\sqrt5)^2}$
use $\#1$ of  this
Or set $\displaystyle 2u+1=\sqrt5\sec\theta$
A: Hint: Obserse that $u^3 + 2u^2 -1$ has $u+1$ as a factor.
A: Hint: $$u^3+2u^2-1=(u+1)(u^2+u-1).$$  Completing the square $$ u^2+u-1= u^2+u\frac{1}{4}-\frac{1}{4}-1=(u+1/2)^2-5/4.$$ 
Now use partial fraction
\begin{align}
\frac{1}{(u+1)((u+1/2)^2-5/4)}=\frac{A}{u+1}+\frac{Bu+C}{( u+1/2)^2-5/4}
\end{align}
We find that $ A=-1, B=1, C=0$, so that 
\begin{align}
I&=\int { \frac{-1}{u+1} du}+\int{ \frac{u}{(u+1/2)^2-5/4}du}
\\
&=-\ln ( u+1 )+J
\end{align}
now to find $ J$, since $$denominator= \frac{5}{4}\left({  (\frac {2}{\sqrt 5}  (u+1/2))^2-1 }\right)$$ we assume $(\frac{2}{\sqrt5}(u+1/2))= w$ to get
\begin{align}
J=constant \int\frac{\sqrt5 w-1}{w^2-1}dw
\end{align}
Finally, setting $ w=\sec\theta $ and proceed $\cdots $
