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

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

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

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

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

Solution steps

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

Factor

ln (x) - ln (1 + x) : - (- ln (x) + ln (1 + x))

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

Take the constant out:

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

= - \int \frac{- ln ( x ) + ln ( 1 + 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 (x) + ln (1 + x)

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

Simplify

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

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

Add a constant to the solution

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

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