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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u}{( u + 1 ) ^{2}} d u

= \int \frac{e ^{u} u}{( u + 1 ) ^{2}} d u

Apply Integration By Parts

: - \frac{e ^{u} u}{u + 1} - \int - e^{u} d u

= - \frac{e ^{u} u}{u + 1} - \int - e^{u} d u

\int - e^{u} d u = - e^{u}

= - \frac{e ^{u} u}{u + 1} - (- e^{u})

Substitute back

u = ln (x)

= - \frac{e ^{l n (x)} ln ( x )}{ln ( x ) + 1} - (- e^{l n (x)})

Simplify

- \frac{e ^{l n (x)} ln ( x )}{ln ( x ) + 1} - (- e^{l n (x)}) : - \frac{x ln ( x )}{ln ( x ) + 1} + x

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

Add a constant to the solution

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

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