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

The integral of xln(sqrt(x+2)) is 1/8 (2x^2ln(x+2)-8ln(x+2)-x^2+4x+12)+C

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integral of xln(sqrt(x+2))

\int x ln (x + 2) d x

\frac{1}{8} (2 x^{2} ln (x + 2) - 8 ln (x + 2) - x^{2} + 4 x + 12) + C

Solution steps

\int x ln (x + 2) d x

Simplify

ln (x + 2) : \frac{1}{2} ln (x + 2)

= \int \frac{1}{2} x ln (x + 2) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x ln (x + 2) d x

Apply u-substitution

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

= \frac{1}{2} \cdot \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

= \frac{1}{2} (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)

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

Substitute back

u = x + 2

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

Simplify

\frac{1}{2} (ln (x + 2) (\frac{( x + 2 ) ^{2}}{2} - 2 (x + 2)) - \frac{1}{2} (\frac{( x + 2 ) ^{2}}{2} - 4 (x + 2))) : \frac{1}{8} (2 x^{2} ln (x + 2) - 8 ln (x + 2) - x^{2} + 4 x + 12)

= \frac{1}{8} (2 x^{2} ln (x + 2) - 8 ln (x + 2) - x^{2} + 4 x + 12)

Add a constant to the solution

= \frac{1}{8} (2 x^{2} ln (x + 2) - 8 ln (x + 2) - x^{2} + 4 x + 12) + C

\int \frac{1}{t ^{2} - 9} d t

\frac{\partial}{\partial x} ((e^{- 6 x}) cos (y))

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

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

What is the integral of xln(sqrt(x+2)) ?
The integral of xln(sqrt(x+2)) is 1/8 (2x^2ln(x+2)-8ln(x+2)-x^2+4x+12)+C