Mathematics

# Evaluate: $\displaystyle\int {(4x + 2)} \sqrt[3]{{{x^2} + x + 1dx}}$

##### SOLUTION
$I=\int { \left( 4x+2 \right) } \sqrt [ 3 ]{ { x }^{ 2 }+x+1 } dx$

let ${ x }^{ 2 }+x+1=t$

$\Rightarrow(2x+1)dx=dt$

$\Rightarrow I=2\int { \left( 2x+1 \right) } \sqrt [ 3 ]{ { x }^{ 2 }+x+1 } =\int { { t }^{ 1/3 } } dt$

$=2.\cfrac { { t }^{ 1/3+1 } }{ 1/3+1 } +c=2\times \cfrac { { t }^{ 4/3 } }{ 4/3 } +c=\cfrac { 3 }{ 2 } { \left( { x }^{ 2 }+x+1 \right) }^{ 4/3 }+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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