Question

The value of the integral $\displaystyle \int x^{ \frac{1}{3}}\left(1 - \sqrt{x}\right)^{3}dx$ is equal to (where $c$ is the constant of integration)

• A. $6\left(\frac{x^{\frac{4}{3}}}{8} + \frac{3}{11} x^{\frac{11}{6}} + \frac{3}{14} x^{\frac{7}{3}} + \frac{1}{17} x^{\frac{17}{6}}\right)+c$
• B. $6\left(\frac{x^{\frac{4}{3}}}{8} - \frac{3}{11} x^{\frac{11}{6}} + \frac{3}{14} x^{\frac{7}{3}} - \frac{1}{17} x^{\frac{17}{6}}\right)+c$
• C. $2\left(\frac{x^{\frac{4}{3}}}{8} - \frac{3}{11} x^{\frac{4}{6}} - \frac{3}{14} x^{\frac{7}{3}} - \frac{1}{17} x^{\frac{17}{6}}\right)+c$
• D. $2\left(\frac{x^{4}}{8} - \frac{3}{11} x^{11} - \frac{3}{14} x^{7} - \frac{1}{17} x^{17}\right)+c$

Answer 1

Answer: (B)

Let, $I=\displaystyle \int x^{\frac{1}{3}}\left(1 - \sqrt{x}\right)^{3 }dx$
Let, $x=t^{6}$ $\Rightarrow dx=6t^{5}dt$
$I=\displaystyle \int t^{2}\left(1 - t^{3}\right)^{3}6t^{5}dt$
$I=6\displaystyle \int t^{7}\left(1 - 3 t^{3} + 3 t^{6} - t^{9}\right)dt$
$=6\displaystyle \int \left(t^{7} - 3 t^{10} + 3 t^{13} - t^{16}\right)dt$
$I=6\left(\frac{t^{8}}{8} - \frac{3 t^{11}}{11} + \frac{3 t^{14}}{14} - \frac{t^{17}}{17}\right)+c$
$I=6\left(\frac{x^{\frac{4}{3}}}{8} - \frac{3}{11} x^{\frac{11}{6}} + \frac{3}{14} x^{\frac{7}{3}} - \frac{1}{17} x^{\frac{17}{6}}\right)+c$