Question

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

Answer 1

Let \(\frac{x}{(x^2+1)(x-1)} = \frac{AX+B}{(x^2+1)}+\frac{c}{(x-1)}\)

x = (Ax+B)(x-1)+C (x²+1)

x = Ax²-Ax+Bx-B+Cx²+C

Equating the coefficients of x2 , x, and constant term, we obtain
A + C = 0
−A + B = 1
−B + C = 0
On solving these equations, we obtain

A = -\(\frac{1}{2}\), B =\(\frac{1}{2}\) and C=\(\frac{1}{2}\)

From equation (1), we obtain

∴\(\frac{x}{(x^2+1)(x-1)}=\frac{\bigg(-\frac{1}{2}x+\frac{1}{2}\bigg)}{x^2+1}+\frac{\frac{1}{2}}{(x-1)}\)

\(\Rightarrow \int\frac{x}{(x^2+1)(x-1)}=-\frac{1}{2}\int\frac{x}{(x^2+1}dx+\frac{1}{2}\int\frac{1}{x^2+1}dx+\frac{1}{2}\int\frac{1}{x-1}dx\)

\(= -\frac{1}{4}\int\frac{2x}{x^2+1}dx+\frac{1}{2}\tan^{-1}x+\frac{1}{2}\log\mid x-1\mid+C\)

\(Consider \int\frac{2x}{x^2+1}dx, let(x^2+1) = t \Rightarrow 2xdx = dt\)

\(\Rightarrow\int\frac{2x}{x^2+1}dx=\int \frac{dt}{t}=\log\mid t\mid=\log\mid x^2+1\mid\)

∴ \(\int\frac{x}{(x^2+1)(x-1)}=-\frac{1}{4}\log\mid x^2+1 \mid+\frac{1}{2}\tan^{-1}x+\frac{1}{2}\log\mid x-1\mid+C\)

=\(\frac{1}{2}\log\mid x-1\mid-\frac{1}{4}\log\mid x^2+1\mid+\frac{1}{2}\tan^{-1}x+C\)