∫1x+xlogxdx I=∫1x(1+logx)dx Let 1+logx=t 1xdx=dt I=∫dtt I=log t+C I=log|1+logx|+C ...

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Special Integrals - 1

∫1x+xlogxdx

Question

$\int \frac{1}{x + x l o g x} d x$

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Solution

$\int \frac{1}{x + x l o g x} d x$

$I = \int \frac{1}{x (1 + l o g x)} d x$

Let $1 + l o g x = t$
$\frac{1}{x} d x = d t$

$I = \int \frac{d t}{t}$

$I = log t + C$

$I = l o g | 1 + l o g x | + C$

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Similar questions

\int \frac{1}{x + x l o g x} d x

\int \frac{1}{x + x log x} d x .

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$\int \frac{1}{x + x l o g x} d x .$

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\int \frac{1}{x log x} d x

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