What is the integral from 0 to 1 of (ln(x))/(5x) ?

The integral from 0 to 1 of (ln(x))/(5x) is diverges

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integral from 0 to 1 of (ln(x))/(5x)

\int_{0}^{1} \frac{ln ( x )}{5 x} d x

diverges

Solution steps

\int_{0}^{1} \frac{ln ( x )}{5 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int_{0}^{1} \frac{ln ( x )}{x} d x

Apply Integration By Parts

= \frac{1}{5} [ln (x) - \int \frac{1 - ln ( x )}{x} d x]_{0}^{1}

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

= \frac{1}{5} [ln (x) - (- \frac{( 1 - ln ( x ) ) ^{2}}{2})]_{0}^{1}

Simplify

\frac{1}{5} [ln (x) - (- \frac{( 1 - ln ( x ) ) ^{2}}{2})]_{0}^{1} : \frac{1}{5} [\frac{ln ^{2} ( x ) + 1}{2}]_{0}^{1}

= \frac{1}{5} [\frac{ln ^{2} ( x ) + 1}{2}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

y = 2 (x + y) \frac{d y}{d x}

d er i v a t i v e x^{\frac{5}{4}}

\frac{d}{d x} ((sin (x) + cos (x))^{3})

d er i v a t i v e (\frac{1}{2} x)^{3}

\int_{4}^{8} \frac{x}{x ^{2} + 6 x + 13} d x

What is the integral from 0 to 1 of (ln(x))/(5x) ?
The integral from 0 to 1 of (ln(x))/(5x) is diverges