What is the integral from 0 to 1/e of 1/(xln^2(x)) ?

The integral from 0 to 1/e of 1/(xln^2(x)) is 1

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integral from 0 to 1/e of 1/(xln^2(x))

\int_{0}^{\frac{1}{e}} \frac{1}{x ln ^{2} ( x )} d x

1

Solution steps

\int_{0}^{\frac{1}{e}} \frac{1}{x ln ^{2} ( x )} d x

Apply Integration By Parts

: [\frac{1 - ln ^{2} ( x ) \cdot \int - \frac{l n ( x ) + 2}{x l n ^{3} ( x )} d x}{ln ^{2} ( x )}]_{0}^{\frac{1}{e}}

= [\frac{1 - ln ^{2} ( x ) \cdot \int - \frac{l n ( x ) + 2}{x l n ^{3} ( x )} d x}{ln ^{2} ( x )}]_{0}^{\frac{1}{e}}

\int - \frac{ln ( x ) + 2}{x ln ^{3} ( x )} d x = \frac{1}{ln ( x )} + \frac{1}{ln ^{2} ( x )}

= [\frac{1 - ln ^{2} ( x ) ( \frac{1}{l n ( x )} + \frac{1}{l n ^{2} ( x )} )}{ln ^{2} ( x )}]_{0}^{\frac{1}{e}}

\frac{1 - ln ^{2} ( x ) ( \frac{1}{l n ( x )} + \frac{1}{l n ^{2} ( x )} )}{ln ^{2} ( x )} = - \frac{1}{ln ( x )}

= [- \frac{1}{ln ( x )}]_{0}^{\frac{1}{e}}

Compute the boundaries:

1

= 1

\int_{- 1}^{2} (x^{2} + x + 1) d x

\int_{- 2021}^{2021} x^{3} d x

\int_{- 2}^{2} - x^{2} - 4 d x

\int_{1}^{3} (x^{2} - 2 x) d x

\int_{- 5}^{0} 2 - 25 - x^{2} d x

What is the integral from 0 to 1/e of 1/(xln^2(x)) ?
The integral from 0 to 1/e of 1/(xln^2(x)) is 1