Question

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

Answer 1

We have,

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

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

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

\[ \Rightarrow x = A \left( x^2 + 1 \right) + B x^2 + Bx + Cx + C\]

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

\[\text{Equating coefficients of like terms}\]

\[A + B = 0 . . . . . \left( 1 \right)\]

\[B + C = 1 . . . . . \left( 2 \right)\]

\[A + C = 0 . . . . . \left( 3 \right)\]

\[\text{Solving (1), (2) and (3), we get}\]

\[A = - \frac{1}{2}\]

\[B = \frac{1}{2}\]

\[C = \frac{1}{2}\]

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

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

\[\text{Let }x^2 + 1 = t\]

\[ \Rightarrow 2x dx = dt\]

\[ \Rightarrow x dx = \frac{dt}{2}\]

\[ \therefore I = - \frac{1}{2}\int\frac{dx}{x + 1} + \frac{1}{4}\int\frac{dt}{t} + \frac{1}{2}\int\frac{dx}{x^2 + 1^2}\]

\[ = - \frac{1}{2} \log \left| x + 1 \right| + \frac{1}{4} \log \left| t \right| + \frac{1}{2} \tan^{- 1} x + C'\]

\[ = - \frac{1}{2} \log \left| x + 1 \right| + \frac{1}{4} \log \left| x^2 + 1 \right| + \frac{1}{2} \tan^{- 1} x + C'\]