Mathematics

# Evaluate the following integrals:$\displaystyle \int { \cfrac { 1 }{ { x }^{ 2/3 }\sqrt { { x }^{ 2/3 }-4 } } } dx\quad$

##### SOLUTION
$I=\displaystyle\int{\dfrac{dx}{{x}^{\frac{2}{3}}\sqrt{{x}^{\frac{2}{3}}-4}}}$

$=\displaystyle\int{\dfrac{{x}^{-\frac{2}{3}}dx}{\sqrt{{x}^{\frac{2}{3}}-4}}}$
Let $t={x}^{\frac{1}{3}}\Rightarrow\,dt=\dfrac{1}{3}{x}^{\frac{1}{3}-1}dx$

$\Rightarrow\,dt=\dfrac{1}{3}{x}^{\frac{-2}{3}}dx$

$\Rightarrow\,3\,dt={x}^{\frac{-2}{3}}dx$

$\therefore\,I=3\displaystyle\int{\dfrac{dt}{\sqrt{{t}^{2}-{2}^{2}}}}$

We know that $\displaystyle\int{\dfrac{dx}{\sqrt{{x}^{2}-{a}^{2}}}}=\log{\left|x+\sqrt{{x}^{2}-{a}^{2}}\right|}+c$
Replace $x\rightarrow\,t$ and $a\rightarrow\,2$

$=3\log{\left|t+\sqrt{{t}^{2}-{2}^{2}}\right|}+c$

$=3\log{\left|t+\sqrt{{t}^{2}-4}\right|}+c$

$=3\log{\left|{x}^{\frac{1}{3}}+\sqrt{{\left({x}^{\frac{1}{3}}\right)}^{2}-4}\right|}+c$ where $t={x}^{\frac{1}{3}}$

$=3\log{\left|{x}^{\frac{1}{3}}+\sqrt{{{x}^{\frac{2}{3}}-4}}\right|}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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