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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int u ln (u) d u

= \int u ln (u) d u

Apply Integration By Parts

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

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

\int \frac{u}{2} d u = \frac{u ^{2}}{4}

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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