Mathematics

# $\displaystyle \int (e^\sqrt[3]{x}dx)=$

$3(x^{2/3} - 2x^{1/3}+2)\exp(\sqrt[3]{x})+c$

##### SOLUTION
$I= \int { { e }^{ \sqrt [ 3 ]{ x } }dx }$
Put $\sqrt [ 3 ]{ x } =t$
$\displaystyle \dfrac { 1 }{ 3 } { x }^{ -2/3 }dx=dt$
$\displaystyle dx=\dfrac { 3 }{ { t }^{ -2 } } dt$

So, $I=3\int { { { t }^{ 2 }e }^{ t }dt }$
$=3[{ { t }^{ 2 }e }^{ t }-2\int { t{ e }^{ t }dt } ]+c$
$=3{ { t }^{ 2 }e }^{ t }-6[t{ e }^{ t }-\int { { e }^{ t }dt } ]+c$
$=3{ { t }^{ 2 }e }^{ t }-6t{ e }^{ t }+6{ e }^{ t }+c$
$=3(x^{2/3} - 2x^{1/3}+2)e^{\sqrt[3]{x}}+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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