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

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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