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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of ln(2x+1)

\int ln (2 x + 1) d x

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

Solution steps

\int ln (2 x + 1) d x

Apply u-substitution

= \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

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

Apply Integration By Parts

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

\int 1 d u = u

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

Substitute back

u = 2 x + 1

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

Simplify

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

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

Add a constant to the solution

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

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