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

The integral of xIn(1+x) is In((x^2)/2+(x^3)/3)+C

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integral of xIn(1+x)

\int x I n (1 + x)

I n (\frac{x ^{2}}{2} + \frac{x ^{3}}{3}) + C

Solution steps

\int x I n (1 + x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= I n \cdot \int (1 + x) x d x

Expand

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

= I n \cdot \int x + x^{2} 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

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

\int x d x = \frac{x ^{2}}{2}

\int x^{2} d x = \frac{x ^{3}}{3}

= I n (\frac{x ^{2}}{2} + \frac{x ^{3}}{3})

Add a constant to the solution

= I n (\frac{x ^{2}}{2} + \frac{x ^{3}}{3}) + C

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What is the integral of xIn(1+x) ?
The integral of xIn(1+x) is In((x^2)/2+(x^3)/3)+C