Integrate the rational function x (x^ 2 +1 )(x-1)

Question

Integrate the rational function $\frac{x}{\left(x^{2}+1\right)(x-1)}$

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Answer

Let $\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C}{(x-1)}$
$x = (Ax + B)(x - 1) + C (x^2 + 1)$
$\Rightarrow x = Ax^2 - Ax + Bx - B + Cx^2 + C$
$\Rightarrow x = (A+C)x^2 - (A-B)x - (B+C)$
Equating the coefficients of $x^2, x$ and constant term, we get,
$A + C = 0$
$-A + B = 1$
$-B + C = 0$
On solving these equation, we get,
$A=-\frac{1}{2}, B=\frac{1}{2} and C=\frac{1}{2}$
Thus,
$\frac{x}{\left(x^{2}+1\right)(x-1)}=\frac{\left(-\frac{1}{2} x+\frac{1}{2}\right)}{\left(x^{2}+1\right)}+\frac{\frac{1}{2}}{(x-1)}$
$\Rightarrow~~\int \frac{x}{\left(x^{2}+1\right)(x-1)} d x=-\frac{1}{2} \int \frac{x}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x-1} d x$
$= -\frac{1}{4} \int \frac{2 x}{x^{2}+1} d x+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+C$
Now, let us consider, $\int \frac{2 x}{x^{2}+1} d x$ Let $(x^2 + 1) = t$
$2xdx = dt$
Thus,
$\int \frac{x}{\left(x^{2}+1\right)(x-1)} d x=-\frac{1}{4} \log \left|\mathrm{x}^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+C$
$= \frac{1}{2} \log |x-1|-\frac{1}{4} \log \left|x^{2}+1\right|+\frac{1}{2} \tan ^{-1} x+C$

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