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

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

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integral of x(ln(x+1))

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

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

Solution steps

\int x ln (x + 1) d x

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 = x + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d y} (e^{x y})

\int_{0}^{\frac{π}{2}} 3 cos^{5} (x) d x

\frac{d}{d x} (x^{\frac{1}{4}} - 4 e^{x})

sin^{7} (θ)

\frac{6}{s} + 16 (\frac{1}{s ^{2} + 16})

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