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 \cdot 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

3 sin (3 x)

\int \frac{x ^{4} - 6 x ^{3} + 12 x ^{2} + 6}{x ^{3} - 6 x ^{2} + 12 x - 8} d x

\frac{d}{d x} (\frac{( x + 4 ) ( x ^{2} - 4 x + 16 )}{x ^{3}})

5 x sin (x)

\int e^{- x} - 1 d x

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