What is the integral from 1 to 2 of (ln(x))/x ?

The integral from 1 to 2 of (ln(x))/x is (ln^2(2))/2

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integral from 1 to 2 of (ln(x))/x

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

\frac{ln ^{2} ( 2 )}{2}

Decimal

0.24022 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{l n (2)} u d u

= \int_{0}^{l n (2)} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{0}^{l n (2)}

= [\frac{u ^{2}}{2}]_{0}^{l n (2)}

Compute the boundaries:

\frac{ln ^{2} ( 2 )}{2}

= \frac{ln ^{2} ( 2 )}{2}

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

\int \frac{1}{h x + a} d x

\frac{d y}{d x} = csc (x) - y cot (x)

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

\int \frac{sin ( x )}{2 - cos ( x )} d x

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