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Question

Mathematics Question on integral

Integrate the rational function: x(x2+1)(x−1)\frac{x}{(x^2+1)(x-1)}

Answer

Let x(x2+1)(x−1)=AX+B(x2+1)+c(x−1)\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 = -12\frac{1}{2}, B =12\frac{1}{2} and C=12\frac{1}{2}

From equation (1), we obtain

∴x(x2+1)(x−1)=(−12x+12)x2+1+12(x−1)\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)}

⇒∫x(x2+1)(x−1)=−12∫x(x2+1dx+12∫1x2+1dx+12∫1x−1dx\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

=−14∫2xx2+1dx+12tan⁡−1x+12log⁡∣x−1∣+C= -\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∫2xx2+1dx,let(x2+1)=t⇒2xdx=dtConsider \int\frac{2x}{x^2+1}dx, let(x^2+1) = t \Rightarrow 2xdx = dt

⇒∫2xx2+1dx=∫dtt=log⁡∣t∣=log⁡∣x2+1∣\Rightarrow\int\frac{2x}{x^2+1}dx=\int \frac{dt}{t}=\log\mid t\mid=\log\mid x^2+1\mid

∴ ∫x(x2+1)(x−1)=−14log⁡∣x2+1∣+12tan⁡−1x+12log⁡∣x−1∣+C\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

=12log⁡∣x−1∣−14log⁡∣x2+1∣+12tan⁡−1x+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