<s>[INST] 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
 [/INST] You are just a baby step away from the answer...: $\dfrac{1}{(x-1)(x^2+1)} = \dfrac{1}{2}\left(\dfrac{1}{x-1} - \dfrac{x}{x^2+1} - \dfrac{1}{x^2+1}\right)$. Can you continue?
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