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

The integral of (x^2+x)ln(x) is 1/3 x^3ln(x)-(x^3)/9+1/2 x^2ln(x)-(x^2)/4+C

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integral of (x^2+x)ln(x)

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

\frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} + \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} + C

Solution steps

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

Expand

(x^{2} + x) ln (x) : x^{2} ln (x) + x ln (x)

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= \frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} + \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4}

Add a constant to the solution

= \frac{1}{3} x^{3} ln (x) - \frac{x ^{3}}{9} + \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} + C

\int sin^{2} (\frac{π x}{2}) d x

(- 3, - 6), (- 9, 1)

\frac{\partial}{\partial y} (\frac{x + y}{y - x})

n \to 2 lim (\frac{3 n + 2}{n})

\int t 1 + t^{2} d t

What is the integral of (x^2+x)ln(x) ?
The integral of (x^2+x)ln(x) is 1/3 x^3ln(x)-(x^3)/9+1/2 x^2ln(x)-(x^2)/4+C