Mathematics

# Solve: $\displaystyle \int \dfrac{x^{e-1}+e^{x-1}}{x^e+e^x}.dx$ ?

##### SOLUTION
Given the integral,
$\int { \dfrac { { e }^{ x-1 }+{ x }^{ e-1 } }{ { x }^{ e }+{ e }^{ x } } } dx$
Let us assume,
$u={ x }^{ e }+{ e }^{ x }\\ \Rightarrow \dfrac { du }{ dx } ={ e }^{ x }+e{ x }^{ e-1 }\\ \Rightarrow du=({ e }^{ x }+e{ x }^{ e-1 })dx$
Substituting these values in the integral we get,
$\int { \dfrac { { e }^{ x-1 }+{ x }^{ e-1 } }{ { x }^{ e }+{ e }^{ x } } } dx\\ =\int { \dfrac { { e }^{ -1 } }{ u } } du\\ ={ e }^{ -1 }\int { \dfrac { 1 }{ u } } du\\ ={ e }^{ -1 }\ln { (u) } \\ ={ e }^{ -1 }\ln { ({ x }^{ e }+{ e }^{ x }) } \\ \therefore \int { \dfrac { { e }^{ x-1 }+{ x }^{ e-1 } }{ { x }^{ e }+{ e }^{ x } } } dx={ e }^{ -1 }\ln { (\left| { x }^{ e }+{ e }^{ x } \right| ) } +C.$

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

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