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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x+2)ln((x+2)^2)

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

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

Solution steps

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

Simplify

ln ((x + 2)^{2}) : 2 ln (x + 2)

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

Take the constant out:

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

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

Expand

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

= 2 \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

= 2 (\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)

= 2 (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

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

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

Add a constant to the solution

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

\int 2021

\int x (x - 2) (x - 3) (x + 2) (x + 3) d x

x \to - 9 lim (- 8)

\frac{d}{d x} (x \cdot ln (x^{2}))

\frac{\partial}{\partial x} (\frac{x z}{y + z})

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