Question

\[\int\frac{1}{x^4 - 1} dx\]

Answer 1

We have,
\[I = \int\frac{dx}{x^4 - 1}\]
\[ = \int\frac{dx}{\left( x^2 - 1 \right) \left( x^2 + 1 \right)}\]
\[ = \int\frac{dx}{\left( x - 1 \right) \left( x + 1 \right) \left( x^2 + 1 \right)}\]
\[\text{Let }\frac{1}{\left( x - 1 \right) \left( x + 1 \right) \left( x^2 + 1 \right)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 1}\]
\[ \Rightarrow \frac{1}{\left( x - 1 \right) \left( x + 1 \right) \left( x^2 + 1 \right)} = \frac{A\left( x^2 + 1 \right) \left( x + 1 \right) + B\left( x - 1 \right) \left( x^2 + 1 \right) \left( Cx + D \right) \left( x - 1 \right) \left( x + 1 \right)}{\left( x - 1 \right) \left( x + 1 \right) \left( x^2 + 1 \right)}\]
\[ \Rightarrow 1 = A\left( x^2 + 1 \right) \left( x + 1 \right) + B \left( x - 1 \right) \left( x^2 + 1 \right) + \left( Cx + D \right) \left( x^2 - 1 \right)\]
\[ \Rightarrow 1 = A\left( x^3 + x^2 + x + 1 \right) + B\left( x^3 + x - x^2 - 1 \right) + \left( C x^3 - Cx + D x^2 - D \right)\]
\[ \Rightarrow 1 = \left( A + B + C \right) x^3 + x^2 \left( A - B + D \right) + x\left( A + B - C \right) + A - B - D\]
\[\text{Equating the coefficients of like terms} . \]
\[A + B + C = 0 . . . . . \left( 1 \right)\]
\[A - B + D = 0 . . . . . \left( 2 \right)\]
\[A + B - C = 0 . . . . . \left( 3 \right)\]
\[A - B - D = 1 . . . . . \left( 4 \right)\]
\[\text{Solving these four equations we get}\]
\[A = \frac{1}{4}, B = - \frac{1}{4}, C = 0, D = - \frac{1}{2}\]
\[ \therefore \frac{1}{\left( x - 1 \right) \left( x + 1 \right) \left( x^2 + 1 \right)} = \frac{1}{4\left( x - 1 \right)} - \frac{1}{4\left( x + 1 \right)} - \frac{1}{2\left( x^2 + 1 \right)}\]
\[ \Rightarrow I = \frac{1}{4}\int \frac{dx}{x - 1} - \frac{1}{4}\int\frac{dx}{x + 1} - \frac{1}{2}\int\frac{dx}{x^2 + 1}\]
\[ = \frac{1}{4}\log \left( x - 1 \right) - \frac{1}{4}\log \left( x + 1 \right) - \frac{1}{2} \tan^{- 1} \left( x \right) + C'\]
\[ = \frac{1}{4}\log \left| \frac{x - 1}{x + 1} \right| - \frac{1}{2} \tan^{- 1} \left( x \right) + C'\]