Question

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

Answer 1

\[\int\frac{\left( x^2 + 1 \right)}{x \left( x^2 - 1 \right)}dx\]
\[ = \int\frac{\left( x^2 + 1 \right)}{x \left( x - 1 \right) \left( x + 1 \right)}dx\]
\[\text{Let }\frac{x^2 + 1}{x \left( x - 1 \right) \left( x + 1 \right)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}\]
\[ \Rightarrow \frac{x^2 + 1}{x \left( x - 1 \right) \left( x + 1 \right)} = \frac{A \left( x - 1 \right) \left( x + 1 \right) + B \left( x \right) \left( x + 1 \right) + C \left( x \right) \left( x - 1 \right)}{x \left( x - 1 \right) \left( x + 1 \right)}\]
\[ \Rightarrow x^2 + 1 = A \left( x - 1 \right) \left( x + 1 \right) + B \left( x \right) \left( x + 1 \right) + C \left( x \right) \left( x - 1 \right) .............(1)\]
\[\text{Putting }x - 1 = 0\text{ or }x = 1\text{ in eq. (1)}\]
\[ \Rightarrow 1 + 1 = A \times 0 + B \left( 1 \right) \left( 1 + 1 \right) + C \times 0\]
\[ \Rightarrow B = 1\]
\[\text{Putting x = 0 in eq. (1)}\]
\[ \Rightarrow 0 + 1 = A \left( 0 - 1 \right) \left( 0 + 1 \right)\]
\[ \Rightarrow A = - 1\]
\[\text{Putting }x + 1 = 0\text{ or }x = - 1\text{ in eq. (1)}\]
\[ \Rightarrow \left( - 1 \right)^2 + 1 = A \times 0 + B \times 0 + C\left( - 1 \right) \left( - 1 - 1 \right)\]
\[ \Rightarrow 2 = C \times 2\]
\[ \Rightarrow C = 1\]
\[ \therefore \frac{x^2 + 1}{x \left( x^2 - 1 \right)} = \frac{- 1}{x} + \frac{1}{x - 1} + \frac{1}{x + 1}\]
\[ \Rightarrow \int\frac{\left( x^2 + 1 \right)}{x \left( x^2 - 1 \right)}dx = - \int\frac{1}{x}dx + \int\frac{1}{x - 1}dx + \int\frac{1}{x + 1}dx\]
\[ = - \ln \left| x \right| + \ln \left| x - 1 \right| + \ln \left| x + 1 \right| + C\]
\[ = - \ln \left| x \right| + \ln \left| x^2 - 1 \right| + C\]
\[ = \ln \left| \frac{x^2 - 1}{x} \right| + C\]