What is the integral of ((y)ln(1+y)) ?

The integral of ((y)ln(1+y)) is 1/2 ln(1+y)(1+y)^2-ln(1+y)-yln(1+y)-1/4 (1+y)^2+1+y+C

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integral of ((y)ln(1+y))

\int ((y) ln (1 + y)) d y

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

Solution steps

\int y ln (1 + y) d y

Apply u-substitution

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

= \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 + y

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

Simplify

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

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

Add a constant to the solution

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

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