Question

Find the integral of the given function w.r.t x

$y=\frac{1}{\sqrt[3]{3x^2}}$

• A. $3\sqrt[3]{3x}$
• B. $\frac{-2}{3}{x}^{-2/3}$
• C. $3\sqrt[3]{3x}+c$
• D. none of these

Answer 1

The correct option is C

$3\sqrt[3]{3x}+c$

$y=\frac{1}{\sqrt[3]{3x^2}}$

The above can be written as:

$y=x^{-2/3}$

now Integration of above w.r.t. x = $\int (x{)}^{-2/3}dx$

We know $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$

$\Rightarrow \int x^{-2/3}dx=\frac{x^{(\frac{-2}{3}+1)}}{\frac{-2}{3}+1}+c$

$\Rightarrow \frac{x^{\frac{1}{3}}}{\frac{1}{3}}+c=3\sqrt[3]{3x}+c$

I hope you remmember why 'c'

Integral also known as antiderivate of a function when differentiated gives back the same function.

So$\Rightarrow \frac{d\left(3\sqrt[3]{3x}\right)}{dx}=\frac{1}{\sqrt[3]{3x^2}}$