Question

$I=\int \frac{(x+x^{\frac{2}{3}}+x^{\frac{1}{6}})}{x(1+x^{\frac{1}{3}})}dx$ is equal to
(where $c$ is constant of integration)

• A. $\frac{3}{2}{x}^{\frac{2}{3}}+6{\tan }^{-1}\left(x^{\frac{1}{6}}\right)+c$
• B. $\frac{3}{2}{x}^{\frac{2}{3}}-6{\tan }^{-1}\left(x^{\frac{1}{6}}\right)+c$
• C. $\frac{3}{2}{x}^{\frac{2}{3}}+{\tan }^{-1}\left(x^{\frac{1}{6}}\right)+c$
• D. None of the above

Answer 1

The correct option is A $\frac{3}{2}{x}^{\frac{2}{3}}+6{\tan }^{-1}\left(x^{\frac{1}{6}}\right)+c$

Substituting $x=p^6,dx=6p^{5}dp,$ we have
$I=\int \frac{6p^5(p^6+p^4+p)}{p^6(1+p^2)}dp=\int \frac{6(p^5+p^3+1)}{(p^2+1)}dp=\int \frac{6\left\{p^3\right(p^2+1)+1\}}{(p^2+1)}dp$
$=\int 6p^{3}dp+\int \left(\frac{6}{p^2+1}\right)dp=\frac{6p^4}{4}+6{\tan }^{-1}p+c=\frac{3}{2}{x}^{\frac{2}{3}}+6{\tan }^{-1}\left(x^{\frac{1}{6}}\right)+c$