Question

\[\int\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} dx\]

Answer 1

We have,
\[I = \int\frac{\left( 5 x^2 - 1 \right) dx}{x \left( x - 1 \right) \left( x + 1 \right)}\]

\[\text{Let }\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}\]

\[ \Rightarrow \frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} = \frac{A \left( x^2 - 1 \right) + Bx \cdot \left( x + 1 \right) + C \cdot x \cdot \left( x - 1 \right)}{x \left( x - 1 \right) \left( x + 1 \right)}\]

\[ \Rightarrow 5 x^2 - 1 = A \left( x^2 - 1 \right) + B \cdot x \left( x + 1 \right) + C \cdot x \cdot \left( x - 1 \right)\]

Putting x = 1

\[ \Rightarrow 5 - 1 = A \times 0 + B \left( 1 \right) \left( 1 + 1 \right) + C \times 0\]

\[ \Rightarrow 4 = B \left( 2 \right)\]

\[ \Rightarrow B = 2\]

Putting x = 0

\[ \Rightarrow 5 \times 0 - 1 = A \left( 0 - 1 \right) + B \times 0 + C \times 0\]

\[ \Rightarrow - 1 = A \left( - 1 \right)\]

\[ \Rightarrow A = 1\]

Putting x + 1 = 0

\[x = - 1\]

\[5 - 1 = A \times 0 + B \times 0 + C \left( - 1 \right) \left( - 2 \right)\]

\[ \Rightarrow C = 2\]

\[ \therefore I = \int\frac{dx}{x} + 2\int\frac{dx}{x - 1} + 2\int\frac{dx}{x + 1}\]

\[ = \log \left| x \right| + 2 \log \left| x - 1 \right| + 2 \log \left| x + 1 \right| + C\]

\[ = \log \left| x \right| + 2 \left\{ \log \left| x^2 - 1 \right| \right\} + C\]

\[ = \log \left| x \left( x^2 - 1 \right)^2 \right| + C\]