Mathematics

# $\displaystyle \int{\dfrac{x^{3}-1}{x^{3}+x}dx}$ equal to

$x-\log x+\dfrac{1}{2}\log(x^{2}+1)-\tan^{-1}x+c$

##### SOLUTION
$I=\int { \dfrac { { { x^{ 3 } }-1 } }{ { { x^{ 3 } }+x } } dx } \\ I=\int { \dfrac { { \left( { x-1 } \right) \left( { { x^{ 2 } }+x+1 } \right) } }{ { x\left( { { x^{ 2 } }+1 } \right) } } dx } \\ I=\int { \dfrac { { x-1 } }{ x } +\int { \dfrac { { x-1 } }{ { { x^{ 2 } }+1 } } dx } }$
$I =\int 1 dx-\int \dfrac{1}{x} dx+\int { \dfrac { { x-1 } }{ { { x^{ 2 } }+1 } } dx }$
$I=x-\ln { x } +\int { \dfrac { { x } }{ { { x^{ 2 } }+1 } } dx } -\int \dfrac{1}{x^2+1}dx$
$I =x-{ \log _{ e } }x+\dfrac { 1 }{ 2 } { \log _{ e } }\left( { { x^{ 2 } }+1 } \right) -{ \tan ^{ -1 } }x+C$

Hence, this is the answer.

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

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