What is the integral of (ln(1+x)-ln(x))/(x(x+1)) ?

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

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integral of (ln(1+x)-ln(x))/(x(x+1))

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

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

Solution steps

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

Apply u-substitution

: \int - u d u

= \int - u d u

Take the constant out:

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

= - \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{u ^{2}}{2}

Substitute back

u = ln (1 + x) - ln (x)

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

Add a constant to the solution

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

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