Question

If $\int \frac{x+\sqrt[3]{3x^2}+\sqrt[6]{6x}}{x(1+\sqrt[3]{3x})}dx=k_1{x}^{\frac{1}{k}}+k_2{\tan }^{-1}\sqrt[m]{mx}+C$, then $2k_1+k_2+m$ equals

• A. $15$
• B. $12$
• C. $\frac{3}{5}$
• D. $\frac{15}{2}$

Answer 1

The correct option is A $15$

Given, $\int \frac{x+\sqrt[3]{3x^2}+\sqrt[6]{6x}}{x(1+\sqrt[3]{3x})}dx=k_1{x}^{\frac{1}{k}}+k_2{\tan }^{-1}\sqrt[m]{mx}+C$ ....(1)
Let $I=\int \frac{x+{\left(x^2\right)}^{\frac{1}{3}}+{\left(x\right)}^{\frac{1}{6}}}{x(1+{\left(x\right)}^{\frac{1}{3}})}dx$
Here L.C.M. of 3 & 6 is 6 so putting $x=t^6$
$\Rightarrow dx=6x^{5}dt$
$\therefore I=6\int \frac{(t^6+t^4+t)}{t^6(1+t^2)}{t}^{5}dt$
$=6\int \frac{t^5+t^3+1}{1+t^2}dt$
$=6\int t^{3}dt+6\int \frac{1}{1+t^2}dt$
$=\frac{3}{2}{t}^4+6{\tan }^{-1}t+C$
$I=\frac{3}{2}{x}^{2/3}+6{\tan }^{-1}{\left(x\right)}^{1/6}+C$
Comparing with (1), we get
$k_1=\frac{3}{2},k_2=6,m=6$
$\therefore 2k_1+k_2+m=3+6+6=15$