Question

\[\int\frac{x^4}{\left( x - 1 \right) \left( x^2 + 1 \right)} dx\]

Answer 1

We have,
\[I = \int \frac{x^4 dx}{\left( x - 1 \right) \left( x^2 + 1 \right)}\]
\[ = \int \left[ \frac{x^4 - 1 + 1}{\left( x - 1 \right) \left( x^2 + 1 \right)} \right]dx\]
\[ = \int \frac{\left( x^4 - 1 \right)dx}{\left( x - 1 \right) \left( x^2 + 1 \right)} + \int\frac{dx}{\left( x - 1 \right) \left( x^2 + 1 \right)}\]
\[ = \int\frac{\left( x^2 - 1 \right) \left( x^2 + 1 \right) dx}{\left( x - 1 \right) \left( x^2 + 1 \right)} + \int\frac{dx}{\left( x - 1 \right) \left( x^2 + 1 \right)}\]
\[ = \int \frac{\left( x - 1 \right) \left( x + 1 \right)dx}{\left( x - 1 \right)} + \int\frac{dx}{\left( x - 1 \right) \left( x^2 + 1 \right)}\]
\[ = \int\left( x + 1 \right)dx + \int\frac{dx}{\left( x - 1 \right) \left( x^2 + 1 \right)} ................\left( 1 \right)\]
\[\text{Let }\frac{1}{\left( x - 1 \right) \left( x^2 + 1 \right)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}\]
\[ \Rightarrow \frac{1}{\left( x - 1 \right) \left( x^2 + 1 \right)} = \frac{A\left( x^2 + 1 \right) + \left( Bx + C \right) \left( x - 1 \right)}{\left( x - 1 \right) \left( x^2 + 1 \right)}\]
\[ \Rightarrow 1 = A x^2 + A + B x^2 - Bx + Cx - C\]
\[ \Rightarrow 1 = \left( A + B \right) x^2 + \left( C - B \right)x + A - C\]
\[\text{Equating coefficients of like terms}\]
\[A + B = 0 . . . . . \left( 1 \right)\]
\[C - B = 0 . . . . . \left( 2 \right)\]
\[A - C = 1 . . . . . \left( 3 \right)\]
\[\text{Solving (1), (2) and (3), we get}\]
\[B = - \frac{1}{2}, A = \frac{1}{2}, C = - \frac{1}{2}\]
\[ \therefore \frac{1}{\left( x - 1 \right) \left( x^2 + 1 \right)} = \frac{1}{2\left( x - 1 \right)} + \frac{- \frac{x}{2} - \frac{1}{2}}{x^2 + 1}\]
\[ \Rightarrow \frac{1}{\left( x - 1 \right) \left( x^2 + 1 \right)} = \frac{1}{2\left( x - 1 \right)} - \frac{1}{2}\left( \frac{x}{x^2 + 1} \right) - \frac{1}{2\left( x^2 + 1 \right)} ............. \left( 2 \right)\]
\[\text{From (1) and (2)}\]
\[I = \int\left( x + 1 \right)dx + \frac{1}{2}\int\frac{dx}{x - 1} - \frac{1}{2}\int\frac{x dx}{x^2 + 1} - \frac{1}{2}\int\frac{dx}{x^2 + 1}\]
\[\text{Putting }x^2 + 1 = t\]
\[ \Rightarrow 2x dx = dt\]
\[ \Rightarrow x dx = \frac{dt}{2}\]
\[ \therefore I = \int\left( x + 1 \right)dx + \frac{1}{2}\int\frac{dx}{x - 1} - \frac{1}{4}\int\frac{dt}{t} - \frac{1}{2}\int\frac{dx}{x^2 + 1}\]
\[ = \frac{x^2}{2} + x + \frac{1}{2}\log \left| x - 1 \right| - \frac{1}{4}\log \left| t \right| - \frac{1}{2} \tan^{- 1} x + C\]
\[ = \frac{x^2}{2} + x + \frac{1}{2}\log \left| x - 1 \right| - \frac{1}{4}\log \left| x^2 + 1 \right| - \frac{1}{2} \tan^{- 1} \left( x \right) + C\]