Question

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

• A. $\frac{3}{2}{x}^{\frac{2}{3}}+6ta{n}^{-1}{x}^{\frac{1}{6}}+c$
• B. $\frac{3}{2}{x}^{\frac{2}{3}}-6ta{n}^{-1}{x}^{\frac{1}{6}}+c$
• C. $-\frac{3}{2}{x}^{\frac{2}{3}}+6ta{n}^{-1}{x}^{\frac{1}{6}}+c$
• D. $-\frac{3}{2}{x}^{\frac{2}{3}}-6ta{n}^{-1}{x}^{\frac{1}{6}}+c$

Answer 1

The correct option is A $\frac{3}{2}{x}^{\frac{2}{3}}+6ta{n}^{-1}{x}^{\frac{1}{6}}+c$

Consider, $I=\int \frac{x+x^{\frac{2}{3}}+x^{\frac{1}{6}}}{x(1+x^{\frac{1}{3}})}dx$

put $x=t^6$

$=dx=6t^{5}dt$

$=\int \frac{t^6+t^4+t}{t^6(1+t^2)}6t^{5}dt$

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

$6\left[\int (t^3+\frac{1}{t^2+1})dt\right]$

$=6[\frac{t^4}{4}+{\tan }^{-1}t]+C$

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