What is the integral from 0 to 1 of 1/(x^{0.5)ln(x)} ?

The integral from 0 to 1 of 1/(x^{0.5)ln(x)} is \Ei(ln(1))

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integral from 0 to 1 of 1/(x^{0.5)ln(x)}

\int_{0}^{1} \frac{1}{x ^{0.5} ln ( x )} d x

Ei (ln (1))

Solution steps

\int_{0}^{1} \frac{1}{x ^{0.5} ln ( x )} d x

Apply u-substitution

: \int_{0}^{1} \frac{1}{ln ( u )} d u

= \int_{0}^{1} \frac{1}{ln ( u )} d u

Apply Integration By Parts

: [\frac{u}{ln ( u )} - \int - \frac{1}{ln ^{2} ( u )} d u]_{0}^{1}

= [\frac{u}{ln ( u )} - \int - \frac{1}{ln ^{2} ( u )} d u]_{0}^{1}

\int - \frac{1}{ln ^{2} ( u )} d u = \frac{u}{ln ( u )} - Ei (ln (u))

= [\frac{u}{ln ( u )} - (\frac{u}{ln ( u )} - Ei (ln (u)))]_{0}^{1}

Simplify

= [Ei (ln (u))]_{0}^{1}

Compute the boundaries:

Ei (ln (1))

= Ei (ln (1))

f (x) = tan (x), at x = π

\frac{d}{d x} (3^{2})

\int (\frac{5 x ^{3}}{2 x ^{2}}) d x

n = 0 \sum \infty \frac{1}{n + 8}

3 x^{3} - 5 x^{2} + 5 x - 14

What is the integral from 0 to 1 of 1/(x^{0.5)ln(x)} ?
The integral from 0 to 1 of 1/(x^{0.5)ln(x)} is \Ei(ln(1))