Question

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

Answer 1

\[\int\left( \frac{x^2 + 1}{x^2 - 1} \right)dx\]
\[ = \int\left( \frac{x^2 - 1 + 2}{x^2 - 1} \right)dx\]
\[ = \int dx + 2\int\frac{1}{x^2 - 1^2}dx\]
\[ = \int dx + 2\int\frac{1}{\left( x - 1 \right) \left( x + 1 \right)}dx . . . \left( 1 \right)\]
\[ \therefore \frac{1}{\left( x - 1 \right)\left( x + 1 \right)} = \frac{A}{x - 1} + \frac{B}{x + 1}\]
\[ \Rightarrow \frac{1}{\left( x - 1 \right) \left( x + 1 \right)} = \frac{A \left( x + 1 \right) + B\left( x - 1 \right)}{\left( x - 1 \right) \left( x + 1 \right)}\]
\[ \Rightarrow 1 = A \left( x + 1 \right) B \left( x - 1 \right) ..........(2)\]
\[\text{Putting }x + 1 = 0\text{ or }x = - 1\text{ in eq. (2)}\]
\[ \Rightarrow 1 = A \times 0 + B \left( - 1 - 1 \right)\]
\[ \Rightarrow B = \frac{- 1}{2}\]
\[\text{Putting }x - 1 = 0\text{ or }x = 1\text{ in eq (2)}\]
\[ \Rightarrow 1 = A \left( 1 + 1 \right) + B \times 0\]
\[ \Rightarrow A = \frac{1}{2}\]
\[ \therefore \frac{1}{\left( x - 1 \right)\left( x + 1 \right)} = \frac{1}{2\left( x - 1 \right)} - \frac{1}{2\left( x + 1 \right)} ..........(3)\]
From eq. (1) and (3)
\[\int\left( \frac{x^2 + 1}{x^2 - 1} \right)dx = \int dx + 2\int\left[ \frac{1}{2 \left( x - 1 \right)} - \frac{1}{2 \left( x + 1 \right)} \right]dx\]
\[ = \int dx + \int\frac{dx}{x - 1} - \int\frac{dx}{x + 1}\]
\[ = x + \ln \left| x - 1 \right| = - \ln \left| x + 1 \right| + C\]
\[ = x + \ln \left| \frac{x - 1}{x + 1} \right| + C\]