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

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

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

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

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

Solution steps

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

Apply Integration By Parts

: - \frac{1 - x}{1 + x} - \int \frac{1}{1 + x} d x

= - \frac{1 - x}{1 + x} - \int \frac{1}{1 + x} d x

\int \frac{1}{1 + x} d x = ln ∣ 1 + x ∣

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

Add a constant to the solution

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

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