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

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 $\displaystyle 2u+1=\sqrt5\sec\theta$
