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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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