Question

\[\int\frac{x^3}{\left( 1 + x^2 \right)^2} \text{ dx }\]

Answer 1

\[\text{ Let  I } = \int\frac{x^3}{\left( 1 + x^2 \right)^2}\text{ dx }\]
\[ = \int\frac{x^2 \times x}{\left( 1 + x^2 \right)^2}\text{ dx }\]
\[\text{  Putting 1 + x}^2 = t \]
\[ \Rightarrow x^2 = t - 1\]
\[ \Rightarrow 2x\text{ dx } = dt\]
\[ \Rightarrow \text{  x dx }= \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int\frac{\left( t - 1 \right)}{t^2}dt\]
\[ = \frac{1}{2}\int\left( \frac{1}{t} - \frac{1}{t^2} \right)\text{ dt }\]
\[ = \frac{1}{2}\int\frac{dt}{t} - \frac{1}{2}\int t^{- 2} \text{ dt }\]
\[ = \frac{1}{2} \text{ ln} \left| t \right| - \frac{1}{2}\left[ \frac{t^{- 2 + 1}}{- 2 + 1} \right] + C\]
\[ = \frac{1}{2} \text{ ln } \left| t \right| + \frac{1}{2t} + C\]
\[ = \frac{1}{2} \text{  ln }\left| 1 + x^2 \right| + \frac{1}{2 \left( 1 + x^2 \right)} + C...... \left( \because t = 1 + x^2 \right)\]