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

The integral of (ln(1+x))/((2-x)^2) is (ln(1+x))/(2-x)-1/3 ln|x+1|+1/3 ln|x-2|+C

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

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

\frac{ln ( 1 + x )}{2 - x} - \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣ + C

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{1}{( 1 + x ) ( 2 - x )} d x = \frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x - 2 ∣

= \frac{ln ( 1 + x )}{2 - x} - (\frac{1}{3} ln ∣ x + 1 ∣ - \frac{1}{3} ln ∣ x - 2 ∣)

Simplify

= \frac{ln ( 1 + x )}{2 - x} - \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{ln ( 1 + x )}{2 - x} - \frac{1}{3} ln ∣ x + 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣ + C

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What is the integral of (ln(1+x))/((2-x)^2) ?
The integral of (ln(1+x))/((2-x)^2) is (ln(1+x))/(2-x)-1/3 ln|x+1|+1/3 ln|x-2|+C