Question

$integrals \: of \: the \: question \: \\ integrate \: it$
​

Answer 1

Answer:

$6( \frac{ {x}^{ \frac{1}{2} } }{ {3 } } - \frac{ {x}^{ \frac{1}{3} } }{2} + {x}^{ \frac{1}{6} } - log( | {x}^{ \frac{1}{6} } + 1| ) ) + c$

Step-by-step explanation:

Let

$x = {t}^{6}$

Then,

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

Now,

$\int \dfrac{1}{ {x}^{ \frac{1}{2} } + {x}^{ \frac{1}{3} } } dx \\ = \int \dfrac{6 {t}^{5} }{ {t}^{3} + {t}^{2} } dt \\ =6 \int \dfrac{ {t}^{3} }{t + 1} dt$

$6\int \dfrac{ {t}^{3} + 1 - 1 }{t + 1} dt \\ 6(\int \dfrac{(t + 1)( {t}^{2} - t + 1)}{t + 1}dt - \int \dfrac{1}{t + 1} dt) \\ = 6(\int ({t}^{2} - t + 1)dt - \int \dfrac{1}{t + 1} dt)$

$6( \frac{ {t}^{3} }{3} - \frac{ {t}^{2} }{2} + t - log( |t + 1| ) ) + c$

Now plugging in the real value:

$6( \frac{ {x}^{ \frac{1}{2} } }{ {3 } } - \frac{ {x}^{ \frac{1}{3} } }{2} + {x}^{ \frac{1}{6} } - log( | {x}^{ \frac{1}{6} } + 1| ) ) + c$