What is the integral of (4x+2)ln(x+1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x+2)ln(x+1)

\int (4 x + 2) ln (x + 1) d x

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

Solution steps

\int (4 x + 2) ln (x + 1) d x

(4 x + 2) ln (x + 1) = 2 (2 x + 1) ln (x + 1)

= \int 2 (2 x + 1) ln (x + 1) d x

Take the constant out:

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

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

Expand

(2 x + 1) ln (x + 1) : 2 x ln (x + 1) + ln (x + 1)

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

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

\int ln (x + 1) d x = x ln (x + 1) + ln (x + 1) - x - 1

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

Simplify

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

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

Add a constant to the solution

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

x \to 1 lim (1 - x^{2})

2^{x}, 0

\int \frac{1}{625 + x ^{2}} d x

\int x^{4} \cdot e^{12 x} d x

2 x y^{^{'}} + y = 2 x

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