What is the integral of (ln(sqrt(1+1/x)))/(x^2+x) ?

The integral of (ln(sqrt(1+1/x)))/(x^2+x) is -1/4 ln^2((x+1)/x)+C

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integral of (ln(sqrt(1+1/x)))/(x^2+x)

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

- \frac{1}{4} ln^{2} (\frac{x + 1}{x}) + C

Solution steps

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

Apply u-substitution

: \int - \frac{ln ( u + 1 )}{2 ( 1 + u )} d u

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

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{ln ( u + 1 )}{1 + u} d u

\frac{ln ( u + 1 )}{1 + u} = ln (u + 1) \frac{1}{1 + u}

= - \frac{1}{2} \cdot \int ln (u + 1) \frac{1}{1 + u} d u

Apply u-substitution

: \int v d v

= - \frac{1}{2} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= - \frac{1}{2} \cdot \frac{v ^{2}}{2}

Substitute back

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

Simplify

- \frac{1}{2} \cdot \frac{ln ^{2} ( 1 + \frac{1}{x} )}{2} : - \frac{1}{4} ln^{2} (\frac{x + 1}{x})

= - \frac{1}{4} ln^{2} (\frac{x + 1}{x})

Add a constant to the solution

= - \frac{1}{4} ln^{2} (\frac{x + 1}{x}) + C

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What is the integral of (ln(sqrt(1+1/x)))/(x^2+x) ?
The integral of (ln(sqrt(1+1/x)))/(x^2+x) is -1/4 ln^2((x+1)/x)+C