Question

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

Answer 1

We have,

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

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

\[ \Rightarrow \frac{1}{\left( x + 1 \right)^2 \left( x^2 + 1 \right)} = \frac{A \left( x + 1 \right) \left( x^2 + 1 \right) + B \left( x^2 + 1 \right) + \left( Cx + D \right) \left( x + 1 \right)^2}{\left( x + 1 \right)^2 \left( x^2 + 1 \right)}\]

\[ \Rightarrow 1 = A \left( x^3 + x + x^2 + 1 \right) + B \left( x^2 + 1 \right) + \left( Cx + D \right)\left( x^2 + 2x + 1 \right)\]

\[ \Rightarrow 1 = A \left( x^3 + x^2 + x + 1 \right) + B \left( x^2 + 1 \right) + C x^3 + 2C x^2 + Cx + D x^2 + 2Dx + D\]

\[ \Rightarrow 1 = \left( A + C \right) x^3 + \left( A + B + 2C + D \right) x^2 + \left( A + C + 2D \right) x + A + B + D\]

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

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

\[A + B + 2C + D = 0 . . . . . \left( 2 \right)\]

\[A + C + 2D = 0 . . . . . \left( 3 \right)\]

\[A + B + D = 1 . . . . . \left( 4 \right)\]

\[A = \frac{1}{2}, B = \frac{1}{2}, C = - \frac{1}{2}\text{ and }D = 0\]

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

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

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

\[ \Rightarrow 2x dx = dt\]

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

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

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

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