Question

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