What is the integral of xln(3+x) ?

The integral of xln(3+x) is ln(3+x)(1/2 (3+x)^2-3(3+x))-1/4 (x^2-6x-27)+C

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integral of xln(3+x)

\int x ln (3 + x) d x

ln (3 + x) (\frac{1}{2} (3 + x)^{2} - 3 (3 + x)) - \frac{1}{4} (x^{2} - 6 x - 27) + C

Solution steps

\int x ln (3 + x) d x

Apply u-substitution

: \int (u - 3) ln (u) d u

= \int (u - 3) ln (u) d u

Apply Integration By Parts

: ln (u) (\frac{u ^{2}}{2} - 3 u) - \int \frac{1}{2} (u - 6) d u

= ln (u) (\frac{u ^{2}}{2} - 3 u) - \int \frac{1}{2} (u - 6) d u

\int \frac{1}{2} (u - 6) d u = \frac{1}{2} (\frac{u ^{2}}{2} - 6 u)

= ln (u) (\frac{u ^{2}}{2} - 3 u) - \frac{1}{2} (\frac{u ^{2}}{2} - 6 u)

Substitute back

u = 3 + x

= ln (3 + x) (\frac{( 3 + x ) ^{2}}{2} - 3 (3 + x)) - \frac{1}{2} (\frac{( 3 + x ) ^{2}}{2} - 6 (3 + x))

Simplify

ln (3 + x) (\frac{( 3 + x ) ^{2}}{2} - 3 (3 + x)) - \frac{1}{2} (\frac{( 3 + x ) ^{2}}{2} - 6 (3 + x)) : ln (3 + x) (\frac{1}{2} (3 + x)^{2} - 3 (3 + x)) - \frac{1}{4} (x^{2} - 6 x - 27)

= ln (3 + x) (\frac{1}{2} (3 + x)^{2} - 3 (3 + x)) - \frac{1}{4} (x^{2} - 6 x - 27)

Add a constant to the solution

= ln (3 + x) (\frac{1}{2} (3 + x)^{2} - 3 (3 + x)) - \frac{1}{4} (x^{2} - 6 x - 27) + C

(- 9 x^{3} + 5 x^{2})^{4}

\int (4 sec (x) tan (x) + 3 sec^{2} (x)) d x

\frac{\partial}{\partial x} (- y^{2})

\int u^{- 3} \frac{1}{2} d u

\frac{d y}{d x} = x e^{- x}

What is the integral of xln(3+x) ?
The integral of xln(3+x) is ln(3+x)(1/2 (3+x)^2-3(3+x))-1/4 (x^2-6x-27)+C