Question

$\int \frac{x+\sqrt[3]{3x^2}+\sqrt[6]{6x}}{x(1+\sqrt[3]{3x})}dx$ is equal to

• A. $\frac{3}{2}{x}^{2/3}+6{\tan }^{-1}{x}^{1/6}+c$
• B. $\frac{3}{2}{x}^{2/3}-6{\tan }^{-1}{x}^{1/6}+c$
• C. $-\frac{3}{2}{x}^{2/3}+6{\tan }^{-1}{x}^{1/6}+c$
• D. None of these

Answer 1

Answer: (A)

$\frac{3}{2}{x}^{2/3}+6{\tan }^{-1}{x}^{1/6}+c$

Put $x=z^6$

$\Rightarrow dx=6z^{5}dz$

$\therefore \int \frac{x+\sqrt[3]{3x^2}+\sqrt[6]{6x}}{x(1+\sqrt[3]{3x})}dx=\int \frac{(z^6+z^4+z)6z^{5}dz}{z^6(1+z^2)}$

$=6\int \frac{z^5+z^3+1}{z^2+1}dz=6\int (z^3+\frac{1}{z^2+1})dz$

$=\frac{3}{2}{z}^4+6{\tan }^{-1}z+c$

$=\frac{3}{2}{x}^{2/3}+6{\tan }^{-1}{x}^{1/6}+c$