<s>[INST] 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
 [/INST] 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 $
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