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

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

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

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

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

Solution steps

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

Add a constant to the solution

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

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