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
 A: 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?
A: Since we aren't permitted to use Partial Fraction, therefore undergo the following method,
$x=\tan \theta \implies dx=\sec^2 \theta\ d\theta$
$\therefore\displaystyle\int \dfrac{dx}{(x-1)\left(x^2+1\right)}=\displaystyle\int \dfrac{\sec^2 \theta\ d\theta}{(\tan \theta-1)\sec^2 \theta}=\displaystyle\int\dfrac{d\theta}{1-\tan \theta }$
$\hat{J}=\displaystyle\int\dfrac{d\theta}{1-\tan \theta },\ \hat{I}=\displaystyle\int\dfrac{\tan \theta}{1-\tan \theta }d\theta$
$$\hat{J}=\displaystyle\int\dfrac{1+\tan \theta}{1-\tan \theta }d\theta-\hat{I}\tag{1} $$
$$\hat{J}=\displaystyle\int d\theta+\hat{I}\tag{2} $$
$$\color{blue}{\therefore \hat{I}=\dfrac{1}{2}\left(\displaystyle\int\dfrac{1+\tan \theta}{1-\tan \theta }d\theta-\displaystyle\int d\theta\right)}$$
A: Hint
Using partial fraction decomposition from the very beginning, we have $$\frac{x^4}{(x-1)(x^2+1)}=\frac{-x-1}{2 \left(x^2+1\right)}+x+\frac{1}{2 (x-1)}+1$$ $$\frac{x^4}{(x-1)(x^2+1)}=\frac{-x}{2 \left(x^2+1\right)}+\frac{-1}{2 \left(x^2+1\right)}+x+\frac{1}{2 (x-1)}+1$$
