Question

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

Answer 1

\[\int\frac{x^3}{\left( x^2 + 1 \right)^3} dx\]
\[ = \int\frac{x^2 . x}{\left( x^2 + 1 \right)^3}dx\]
\[\text{Let }x^2 + 1 = t\]
\[ \Rightarrow x^2 = t - 1\]
\[ \Rightarrow \text{2x dx} = dt\]
\[ \Rightarrow\text{ x dx} = \frac{dt}{2}\]
\[Now, \int\frac{x^2 . x}{\left( x^2 + 1 \right)^3}dx\]
\[ = \frac{1}{2}\int\frac{\left( t - 1 \right)}{t^3}dt\]
\[ = \frac{1}{2}\int\left( \frac{1}{t^2} - \frac{1}{t^3} \right) dt\]
\[ = \frac{1}{2}\int\left( t^{- 2} - t^{- 3} \right)dt\]
\[ = \frac{1}{2}\left[ \frac{t^{- 2 + 1}}{- 2 + 1} - \frac{t^{- 3 + 1}}{- 3 + 1} \right] + C\]
\[ = \frac{1}{2}\left[ - \frac{1}{t} + \frac{1}{2 t^2} \right] + C\]
\[ = \frac{1}{2}\left[ \frac{- 1}{\left( x^2 + 1 \right)} + \frac{1}{2 \left( x^2 + 1 \right)^2} \right] + C\]
\[ = \frac{1}{2}\left[ \frac{- 2 \left( x^2 + 1 \right) + 1}{2 \left( x^2 + 1 \right)^2} \right]\]
\[ = \frac{1}{4}\left[ \frac{- 2 x^2 - 2 + 1}{\left( x^2 + 1 \right)^2} \right] = - \frac{1}{4}\frac{\left( 1 + 2 x^2 \right)}{\left( x^2 + 1 \right)^2} + C\]