Question

\[\int\frac{x^2}{\sqrt{3x + 4}} dx\]

Answer 1

\[\int\frac{x^2 dx}{\sqrt{3x + 4}}\]
\[\text{Let 3x + 4 }= t \]
\[ \Rightarrow x = \frac{t - 4}{3}\]
\[ \Rightarrow 1 = \frac{1}{3} . \frac{dt}{dx}\]
\[ \Rightarrow dx = \frac{dt}{3}\]
\[Now, \int\frac{x^2 dx}{\sqrt{3x + 4}}\]
\[ = \frac{1}{3}\int\frac{\left( \frac{t - 4}{3} \right)^2}{\sqrt{t}}dt\]
\[ = \frac{1}{27}\int\left( \frac{t^2}{\sqrt{t}} - \frac{8t}{\sqrt{t}} + \frac{16}{\sqrt{t}} \right)dt\]

\[ = \frac{1}{27}\int\left( t^\frac{3}{2} - 8 t^\frac{1}{2} + 16 t^{- \frac{1}{2}} \right)dt\]
\[ = \frac{1}{27} \left[ \frac{t^\frac{3}{2} + 1}{\frac{3}{2} + 1} + \frac{8 t^\frac{1}{2} + 1}{\frac{1}{2} + 1} + \frac{16 t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + C\]
\[ = \frac{1}{27} \left[ \frac{2}{5} t^\frac{5}{2} - \frac{8 \times 2}{3} t^\frac{3}{2} + 32 t^\frac{1}{2} \right] + C\]
\[ = \frac{2}{135} \left( t \right)^\frac{5}{2} - \frac{16}{81} t^\frac{3}{2} + \frac{32}{27} t^\frac{1}{2} + C\]
\[ = \frac{2}{135} \left( 3x + 4 \right)^\frac{5}{2} - \frac{16}{81} \left( 3x + 4 \right)^\frac{3}{2} + \frac{32}{27} \left( 3x + 4 \right)^\frac{1}{2} + C\]