What is the integral of x^2ln(x+1) ?

The integral of x^2ln(x+1) is -x^2ln(x+1)-xln(x+1)+1/3 ln(x+1)(x+1)^3-1/9 (x+1)^3+1/2 (x+1)^2-x-1+C

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integral of x^2ln(x+1)

integral

x^{2} ln (x + 1)

- x^{2} ln (x + 1) - x ln (x + 1) + \frac{1}{3} ln (x + 1) (x + 1)^{3} - \frac{1}{9} (x + 1)^{3} + \frac{1}{2} (x + 1)^{2} - x - 1 + C

Solution steps

\int x^{2} ln (x + 1) d x

Apply u-substitution

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

Apply Integration By Parts

= ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - \int \frac{1}{3} (u^{2} - 3 u + 3) d u

\int \frac{1}{3} (u^{2} - 3 u + 3) d u = \frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u

= ln (u) (\frac{u ^{3}}{3} - u^{2} + u) - (\frac{u ^{3}}{9} - \frac{u ^{2}}{2} + u)

Substitute back

u = x + 1

= ln (x + 1) (\frac{( x + 1 ) ^{3}}{3} - (x + 1)^{2} + x + 1) - (\frac{( x + 1 ) ^{3}}{9} - \frac{( x + 1 ) ^{2}}{2} + x + 1)

Simplify

ln (x + 1) (\frac{( x + 1 ) ^{3}}{3} - (x + 1)^{2} + x + 1) - (\frac{( x + 1 ) ^{3}}{9} - \frac{( x + 1 ) ^{2}}{2} + x + 1) : - x^{2} ln (x + 1) - x ln (x + 1) + \frac{1}{3} ln (x + 1) (x + 1)^{3} - \frac{1}{9} (x + 1)^{3} + \frac{1}{2} (x + 1)^{2} - x - 1

= - x^{2} ln (x + 1) - x ln (x + 1) + \frac{1}{3} ln (x + 1) (x + 1)^{3} - \frac{1}{9} (x + 1)^{3} + \frac{1}{2} (x + 1)^{2} - x - 1

Add a constant to the solution

= - x^{2} ln (x + 1) - x ln (x + 1) + \frac{1}{3} ln (x + 1) (x + 1)^{3} - \frac{1}{9} (x + 1)^{3} + \frac{1}{2} (x + 1)^{2} - x - 1 + C

What is the integral of x^2ln(x+1) ?
The integral of x^2ln(x+1) is -x^2ln(x+1)-xln(x+1)+1/3 ln(x+1)(x+1)^3-1/9 (x+1)^3+1/2 (x+1)^2-x-1+C