What is the integral of xln(1-x) ?

The integral of xln(1-x) is 1/2 ln(1-x)(1-x)^2-ln(1-x)+xln(1-x)-1/4 (1-x)^2+1-x+C

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integral of xln(1-x)

\int x ln (1 - x) d x

\frac{1}{2} ln (1 - x) (1 - x)^{2} - ln (1 - x) + x ln (1 - x) - \frac{1}{4} (1 - x)^{2} + 1 - x + C

Solution steps

\int x ln (1 - x) d x

Apply u-substitution

: \int - (- u + 1) ln (u) d u

= \int - (- u + 1) ln (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int (- u + 1) ln (u) d u

Apply Integration By Parts

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

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

\int \frac{1}{2} (- u + 2) d u = - \frac{u ^{2}}{4} + u

= - (ln (u) (- \frac{u ^{2}}{2} + u) - (- \frac{u ^{2}}{4} + u))

Substitute back

u = 1 - x

= - (ln (1 - x) (- \frac{( 1 - x ) ^{2}}{2} + 1 - x) - (- \frac{( 1 - x ) ^{2}}{4} + 1 - x))

Simplify

- (ln (1 - x) (- \frac{( 1 - x ) ^{2}}{2} + 1 - x) - (- \frac{( 1 - x ) ^{2}}{4} + 1 - x)) : \frac{1}{2} ln (1 - x) (1 - x)^{2} - ln (1 - x) + x ln (1 - x) - \frac{1}{4} (1 - x)^{2} + 1 - x

= \frac{1}{2} ln (1 - x) (1 - x)^{2} - ln (1 - x) + x ln (1 - x) - \frac{1}{4} (1 - x)^{2} + 1 - x

Add a constant to the solution

= \frac{1}{2} ln (1 - x) (1 - x)^{2} - ln (1 - x) + x ln (1 - x) - \frac{1}{4} (1 - x)^{2} + 1 - x + C

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What is the integral of xln(1-x) ?
The integral of xln(1-x) is 1/2 ln(1-x)(1-x)^2-ln(1-x)+xln(1-x)-1/4 (1-x)^2+1-x+C