Explanation

(c)
Let $I = \int_{}^{}\frac{dx}{\sqrt{x} + \sqrt[3]{x}}$
Now, let $x = t^{6} \Rightarrow dx = 6t^{5}dt$
$\Rightarrow I = \int_{}^{}{\frac{6t^{5}}{t^{3} + t^{2}}dt = 6\int_{}^{}{\frac{t^{3}}{t + 1}dt}}$
$= 6\int_{}^{}{\left( t^{2} - t + 1 - \frac{1}{t + 1} \right)dt}$
$= 2t^{3} - 3t^{2} + 6t - 6\log{(t + 1) + c}$
$= 2\sqrt{x} - 3\left( \sqrt[3]{x} \right) + 6\left( \sqrt[6]{x} \right) - 6\log{\left( \sqrt[6]{x} + 1 \right) + c}$