Question

If $\int \frac{x+\sqrt[3]{3x^2}+\sqrt[6]{6x}}{x(1+\sqrt[3]{3x})}dx=a{x}^{\frac{2}{3}}+b{\tan }^{-1}\left(\sqrt[6]{6x}\right)+c$ then:

• A. $a=\frac{3}{2}$
• B. $a=\frac{3}{4}$
• C. $a=\frac{-3}{2}$
• D. $a=\frac{-3}{4}$

Answer 1

Answer: (A)

$a=\frac{3}{2}$

$\int \frac{x+x^{2/3}+x^{1/6}}{x(x^{1/3}+1)}dx$

$u=x^{1/6}$

$du=\frac{1}{6}\frac{dx}{x^{5/6}}$

$x^{1/3}=u^2$

$x^{2/3}=u^4$

$x=u^6$

$x=u^6$

$=6\int \frac{u^5+u^3+1}{u^2+1}du$

Polynomial division

$\Rightarrow 6\int (\frac{1}{u^2+1}+u^3)du$

$\Rightarrow 6{\tan }^{-1}\left(u\right)+\frac{{6u}^4}{4}+c$

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