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

The integral of ln(x+1)x 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 ln(x+1)x

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

Apply u-substitution

= \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

\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

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