Question

$\int \: \frac{1}{ {x}^{ \frac{1}{2} }+ {x}^{ \frac{1}{3} } } dx$
Solve this ....​

Answer 1

Answer:

$\int \dfrac{1}{ {x}^{ \frac{1}{2} } + {x}^{ \frac{1}{3} } } dx$

$Let \: x = {t}^{6 } \\ = > dx = 6 {t}^{5} dt$

Putting the value :

$= \int \dfrac{6 {t}^{5} }{ {t}^{3} + {t}^{2} } dt$

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

$= 6 \int \dfrac{ {t}^{3} }{1 + t} dt$

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

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

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

Reversing the value in terms of x :

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

So final answer :

$\boxed{ \red{ \sf{ \bold{6 \{ \frac{ {x}^{ \frac{1}{2} } }{3} - \frac{ {x}^{ \frac{1}{3} } }{2} + {x}^{ \frac{1}{6} } - log( {x}^{ \frac{1}{6} } + 1) \} + c}}}}$

The basic trick for the substitution :

• It has been given x^½ and x^⅓ . So we are taking t^6 because the LCM of 2 and 3 is 6

Additional Information:

• We have performed Chain rule in this sum , such that we try to express dx in terms of dt.

Answer 2

Answer is given in the pic.