What is the integral of 6(x+2)*ln(x+2) ?

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

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integral of 6(x+2)*ln(x+2)

\int 6 (x + 2) \cdot ln (x + 2) d x

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

Solution steps

\int 6 (x + 2) ln (x + 2) d x

Take the constant out:

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

= 6 \cdot \int (x + 2) ln (x + 2) d x

Expand

(x + 2) ln (x + 2) : x ln (x + 2) + 2 ln (x + 2)

= 6 \cdot \int x ln (x + 2) + 2 ln (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

= 6 (\int x ln (x + 2) d x + \int 2 ln (x + 2) d x)

\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)

\int 2 ln (x + 2) d x = 2 (x ln (x + 2) + 2 ln (x + 2) - x - 2)

= 6 (ln (x + 2) (\frac{1}{2} (x + 2)^{2} - 2 (x + 2)) - \frac{1}{4} (x^{2} - 4 x - 12) + 2 (x ln (x + 2) + 2 ln (x + 2) - x - 2))

Simplify

6 (ln (x + 2) (\frac{1}{2} (x + 2)^{2} - 2 (x + 2)) - \frac{1}{4} (x^{2} - 4 x - 12) + 2 (x ln (x + 2) + 2 ln (x + 2) - x - 2)) : 6 (\frac{1}{2} x^{2} ln (x + 2) - \frac{1}{4} x^{2} + 2 x ln (x + 2) - x + 2 ln (x + 2) - 1)

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

Add a constant to the solution

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

\frac{d}{d x} ((3 x + 2)^{2})

\int \frac{4}{y} d y

d er i v a t i v e x^{3} + 1

x \to - \infty lim (3 x + 9 x^{2} + 7 x - 8)

\int 16 x d x

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