Mathematics

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


SOLUTION

We have,

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

LCM of $$2\,and\,3$$ $$=6$$

Let,

$$ x={{t}^{6}} $$

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

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

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

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

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

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

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

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

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

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

$$ =6\left[ \dfrac{{{t}^{3}}}{3} \right]+6t-6\left[ \dfrac{{{t}^{2}}}{2} \right]-6\log \left( t+1 \right)+C $$

Put $$t=\sqrt[6]{x}$$

$$=6\dfrac{\sqrt{x}}{3}+6{{x}^{\dfrac{1}{6}}}-6{{x}^{\dfrac{1}{3}}}-6\log \left( {{x}^{\dfrac{1}{6}}}+1 \right)+C$$

Hence, this is the answer.
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