What is the integral of xln(x+4) ?

The integral of xln(x+4) is ln(x+4)(1/2 (x+4)^2-4(x+4))-1/4 (x^2-8x-48)+C

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integral of xln(x+4)

\int x ln (x + 4) d x

ln (x + 4) (\frac{1}{2} (x + 4)^{2} - 4 (x + 4)) - \frac{1}{4} (x^{2} - 8 x - 48) + C

Solution steps

\int x ln (x + 4) d x

Apply u-substitution

: \int (u - 4) ln (u) d u

= \int (u - 4) ln (u) d u

Apply Integration By Parts

: ln (u) (\frac{u ^{2}}{2} - 4 u) - \int \frac{1}{2} (u - 8) d u

= ln (u) (\frac{u ^{2}}{2} - 4 u) - \int \frac{1}{2} (u - 8) d u

\int \frac{1}{2} (u - 8) d u = \frac{1}{2} (\frac{u ^{2}}{2} - 8 u)

= ln (u) (\frac{u ^{2}}{2} - 4 u) - \frac{1}{2} (\frac{u ^{2}}{2} - 8 u)

Substitute back

u = x + 4

= ln (x + 4) (\frac{( x + 4 ) ^{2}}{2} - 4 (x + 4)) - \frac{1}{2} (\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4))

Simplify

ln (x + 4) (\frac{( x + 4 ) ^{2}}{2} - 4 (x + 4)) - \frac{1}{2} (\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4)) : ln (x + 4) (\frac{1}{2} (x + 4)^{2} - 4 (x + 4)) - \frac{1}{4} (x^{2} - 8 x - 48)

= ln (x + 4) (\frac{1}{2} (x + 4)^{2} - 4 (x + 4)) - \frac{1}{4} (x^{2} - 8 x - 48)

Add a constant to the solution

= ln (x + 4) (\frac{1}{2} (x + 4)^{2} - 4 (x + 4)) - \frac{1}{4} (x^{2} - 8 x - 48) + C

\int ln (y) + 2 x y^{3} d x

- x y + (3 + x) (\frac{d y}{d x}) = 0

\int_{0}^{\frac{1}{2}} 2 arccos (x) d x

\int \frac{20}{x ^{- 1} + 5} d x

\frac{\partial}{\partial x} (2 x y^{- 1})

What is the integral of xln(x+4) ?
The integral of xln(x+4) is ln(x+4)(1/2 (x+4)^2-4(x+4))-1/4 (x^2-8x-48)+C