Question

\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]

Answer 1

\[Let\ I = \int_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} d x . \]
\[Let\ 2x + 1 = u\]
\[ \Rightarrow x = \frac{u - 1}{2}\]
\[ \Rightarrow dx = \frac{du}{2}\]
\[ \therefore I = \int\frac{\left( \frac{u - 1}{2} \right)^2 + \frac{u - 1}{2}}{\sqrt{u}} \frac{du}{2}\]
\[ \Rightarrow I = \frac{1}{8}\int\frac{u^2 + 1 - 2u + 2u - 2}{\sqrt{u}} du\]
\[ = \frac{1}{8}\int\frac{\left( u^2 - 1 \right)}{\sqrt{u}} du\]
\[ = \frac{1}{8}\int\left( u^\frac{3}{2} - u^{- \frac{1}{2}} \right) du\]
\[ = \frac{1}{8}\left[ \frac{2 u^\frac{5}{2}}{5} - \frac{2 u^\frac{1}{2}}{1} \right]\]
\[ = \frac{1}{8} \left[ \frac{2 \left( 2x + 1 \right)^\frac{5}{2}}{5} - \frac{2 \left( 2x + 1 \right)^\frac{1}{2}}{1} \right]_1^4 \]
\[ = \frac{1}{8}\left[ \frac{2}{5} \times 243 - 6 - \frac{2}{5} \times 9\sqrt{3} + 2\sqrt{3} \right]\]
\[ \Rightarrow I = \frac{1}{8}\left[ \frac{456}{5} - \frac{8\sqrt{3}}{5} \right]\]
\[ \Rightarrow I = \frac{57 - \sqrt{3}}{5}\]