Evaluation of $\int \frac{x^4}{(x-1)(x^2+1)}dx$ Evaluation of $\displaystyle \int \frac{x^4}{(x-1)(x^2+1)}dx$
$\bf{My\; Try::}$ Let $$\displaystyle I = \int\frac{x^4}{(x-1)(x^2+1)}dx = \int \frac{(x^4-1)+1}{(x-1)(x^2+1)}dx = \int\frac{(x-1)\cdot (x+1)\cdot (x^2+1)}{(x-1)(x^2+1)}+\int\frac{1}{(x-1)(x^2+1)}dx$$
So $\displaystyle I = \int (x+1)dx+J\;\,\;,$ Where $\displaystyle J = \int\frac{1}{(x-1)(x^2+1)}dx$
Now can we solve $J$ without using Partial fraction.
If yes then plz explain me, Thanks

Hint
Using partial fraction decomposition from the very beginning, we have $$\frac{x^4}{(x-1)(x^2+1)}=\frac{-x-1}{2 \left(x^2+1\right)}+x+\frac{1}{2 (x-1)}+1$$ $$\frac{x^4}{(x-1)(x^2+1)}=\frac{-x}{2 \left(x^2+1\right)}+\frac{-1}{2 \left(x^2+1\right)}+x+\frac{1}{2 (x-1)}+1$$
