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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4ln(2x+1)

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

2 (2 x ln (2 x + 1) + ln (2 x + 1) - 2 x - 1) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int ln (u) \frac{1}{2} d u

= 4 \cdot \int ln (u) \frac{1}{2} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

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

\int 1 d u = u

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

Substitute back

u = 2 x + 1

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

Simplify

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

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

Add a constant to the solution

= 2 (2 x ln (2 x + 1) + ln (2 x + 1) - 2 x - 1) + C

\frac{d}{d x} ((\frac{cos ( x )}{1 + sin ( x )})^{2})

\int \frac{3}{2} x^{3} d x

3 x^{2} ln (x), 12 ln (x)

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

\frac{\partial}{\partial x} (6 y^{2} - 4 y^{3} - 3 x^{2} + 6 x y)

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