What is the integral of xln(x+2) ?

The integral of xln(x+2) is ln(x+2)(1/2 (x+2)^2-2(x+2))-1/4 (x^2-4x-12)+C

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integral of xln(x+2)

\int x ln (x + 2) d x

ln (x + 2) (\frac{1}{2} (x + 2)^{2} - 2 (x + 2)) - \frac{1}{4} (x^{2} - 4 x - 12) + C

Solution steps

\int x ln (x + 2) d x

Apply u-substitution

: \int (u - 2) ln (u) d u

= \int (u - 2) ln (u) d u

Apply Integration By Parts

: ln (u) (\frac{u ^{2}}{2} - 2 u) - \int \frac{1}{2} (u - 4) d u

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

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

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

Substitute back

u = x + 2

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

Simplify

ln (x + 2) (\frac{( x + 2 ) ^{2}}{2} - 2 (x + 2)) - \frac{1}{2} (\frac{( x + 2 ) ^{2}}{2} - 4 (x + 2)) : ln (x + 2) (\frac{1}{2} (x + 2)^{2} - 2 (x + 2)) - \frac{1}{4} (x^{2} - 4 x - 12)

= ln (x + 2) (\frac{1}{2} (x + 2)^{2} - 2 (x + 2)) - \frac{1}{4} (x^{2} - 4 x - 12)

Add a constant to the solution

= ln (x + 2) (\frac{1}{2} (x + 2)^{2} - 2 (x + 2)) - \frac{1}{4} (x^{2} - 4 x - 12) + C

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9 x^{2}

\frac{d}{d x} (ln (x^{2} - 5 x + 2))

What is the integral of xln(x+2) ?
The integral of xln(x+2) is ln(x+2)(1/2 (x+2)^2-2(x+2))-1/4 (x^2-4x-12)+C