Question

\[\int\frac{x^5}{\sqrt{1 + x^3}} \text{ dx }\]

Answer 1

\[\text{We have}, \]
\[I = \int\frac{x^5 dx}{\sqrt{1 + x^3}}\]
\[ = \int\frac{x^3 x^2 \text{ dx }}{\sqrt{1 + x^3}}\]
\[\text{ Putting x}^3 = t \]
\[ \Rightarrow 3 x^2 \text{ dx }= dt\]
\[ \Rightarrow x^2 \text{ dx }= \frac{dt}{3}\]
\[ \therefore I = \frac{1}{3}\int\frac{t \text{ dt}}{\sqrt{1 + t}}\]
\[ = \frac{1}{3}\int\left( \frac{1 + t - 1}{\sqrt{1 + t}} \right) dt\]
\[ = \frac{1}{3}\int\left( \sqrt{1 + t} - \frac{1}{\sqrt{1 + t}} \right) dt\]
\[ = \frac{1}{3} \left[ \frac{\left( 1 + t \right)^\frac{3}{2}}{\frac{3}{2}} - \frac{\left( 1 + t \right)^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + C\]
\[ = \frac{2}{9} \left( 1 + t \right)^\frac{3}{2} - \frac{2}{3} \left( 1 + t \right)^\frac{1}{2} + C\]
\[ = \frac{2}{9} \left( 1 + x^3 \right)^\frac{3}{2} - \frac{2}{3} \left( 1 + x^3 \right)^\frac{1}{2} + C\]
\[ = \frac{2}{9} \left( 1 + x^3 \right)^\frac{1}{2} \left( 1 + x^3 - 3 \right) + C\]
\[ = \frac{2}{9} \sqrt{1 + x^3} \left( x^3 - 2 \right) + C\]