Question

Evaluate the following integrals:

\[\int\frac{x^2}{(x - 1) ( x^2 + 1)}dx\]

Answer 1

\[\int\frac{x^2}{(x - 1) ( x^2 + 1)}dx\]

\[\frac{x^2}{(x - 1)( x^2 + 1)} = \frac{A}{(x - 1)} + \frac{Bx + C}{x^2 + 1}\]
\[ = \frac{A\left( x^2 + 1 \right) + \left( Bx + C \right)\left( x - 1 \right)}{(x - 1)( x^2 + 1)}\]
\[ \Rightarrow \frac{x^2}{(x - 1)( x^2 + 1)} = \frac{(A + B) x^2 + (C - B)x + \left( A - C \right)}{(x - 1)( x^2 + 1)}\]

Comparing coefficients, we get

\[A + B = 1; C - B = 0\text{ and }A - C = 0\]
\[\text{Solving these equations, we get}\]
\[A = B = C = \frac{1}{2}\]
\[ \therefore I = \frac{1}{2}\int\frac{1}{(x - 1)}dx + \frac{1}{2}\int\frac{x}{x^2 + 1}dx + \frac{1}{2}\int\frac{1}{x^2 + 1}dx\]
\[ = \frac{1}{2}\ln\left| x - 1 \right| + \frac{1}{4}\ln\left| x^2 + 1 \right| + \frac{1}{2} \tan^{- 1} \left( x \right) + c\]