Question

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

Answer 1

\[\int\frac{x^5 dx}{\sqrt{1 + x^3}}\]
\[ = \int\frac{x^3 . x^2 dx}{\sqrt{1 + x^3}}\]
\[\text{Let 1} + x^3 = t \]
\[ \Rightarrow x^3 = t - 1\]
\[ \Rightarrow 3 x^2 = \frac{dt}{dx}\]
\[ \Rightarrow \text{x^2 dx} = \frac{dt}{3}\]
` Now,∫  {x^3 . x^2     dx}/{\sqrt{1 + x^3}}`
\[ = \frac{1}{3}\int\frac{\left( t - 1 \right)}{\sqrt{t}} dt\]
\[ = \frac{1}{3}\int\left( \sqrt{t} - \frac{1}{\sqrt{t}} \right)dt\]
\[ = \frac{1}{3} \int\left( t^\frac{1}{2} - t^{- \frac{1}{2}} \right)dt\]
\[ = \frac{1}{3}\left[ \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} - \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + C\]
\[ = \frac{1}{3}\left[ \frac{2}{3} t^\frac{3}{2} - 2\sqrt{t} \right] + C\]
\[ = \frac{2}{9} \left( 1 + x^3 \right)^\frac{3}{2} - \frac{2}{3} \left( 1 + x^3 \right)^\frac{1}{2} + C\]