Question

If $I=\int \frac{dx}{\sqrt[3]{x^{\frac{5}{2}}{\left(1+x\right)}^{\frac{7}{2}}}}$ $=kf\left(x\right)+c,$ where $c$ is the integration constant and $f\left(1\right)=\frac{1}{2^{\frac{1}{6}}},$ then the value of $f\left(2\right)$ is equal to

• A. $6{\left(\frac{2}{3}\right)}^{\frac{1}{6}}$
• B. $6{\left(\frac{3}{2}\right)}^{\frac{1}{6}}$
• C. ${\left(\frac{2}{3}\right)}^{\frac{1}{6}}$
• D. ${\left(\frac{2}{3}\right)}^6$

Answer 1

Answer: (C)

The given integral is $I=\int \frac{dx}{{\left(\left[x^6{\left(\frac{1+x}{x}\right)}^{\frac{7}{2}}\right]\right)}^{\frac{1}{3}}}$
$\Rightarrow I=\int \frac{dx}{x^2{\left(1+\frac{1}{x}\right)}^{\frac{7}{6}}}$
Let, $1+\frac{1}{x}=t$
$\Rightarrow -\frac{dx}{x^2}=dt$
$\Rightarrow I=\int -\frac{dt}{t^{\frac{7}{6}}}$
$=6t^{-\frac{1}{6}}+c$
$=6{\left(\frac{x}{x+1}\right)}^{\frac{1}{6}}+c$
$\therefore f\left(x\right)={\left(\frac{x}{x+1}\right)}^{\frac{1}{6}}$
$\Rightarrow f\left(2\right)={\left(\frac{2}{3}\right)}^{\frac{1}{6}}$