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

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

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integral of (x[ln(1+x)+ln(1-x)])/(x^2-1)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u}{2} d u

= \int \frac{u}{2} 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 u d u

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{u ^{2}}{2}

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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