Mathematics

# Solve :$\displaystyle \int \frac{1}{x+x\log x}dx.$

$\displaystyle \log \left ( 1+\log x \right ).$

##### SOLUTION
Let $\displaystyle I=\int \frac { 1 }{ x+x\log x } dx=\int \frac { 1 }{ x\left( 1+\log x \right) } dx$

Put $\displaystyle 1+\log x=t\Rightarrow \frac { 1 }{ x } dx=dt$

Therefore

$\displaystyle I=\int { \frac { dt }{ t } } =\log { t } =\log \left( 1+\log x \right)$

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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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