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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = ln (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{( 1 + t ^{4} )}

(1 + x)^{n}

\frac{1}{1 - t ^{2}}

(x + 1)^{2}

a^{12} y^{4} - a^{4} y^{12}

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