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

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

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integral of x^2ln(1+x)

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

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

Solution steps

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

Apply u-substitution

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

= \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

= 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 = 1 + x

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

Simplify

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

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

Add a constant to the solution

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

\int 4 x cos (4 x^{2}) d x

\frac{d y}{d x} = y - 1

\frac{d y}{d x} = \frac{y ^{3}}{3 x - 2}, y (1) = - 1

\int_{- \infty}^{5} \frac{1}{x ^{2} + 1} d x

\int 1 + cos (θ) d θ

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