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

The integral of (x+2)/(2x-1) is 1/2 x-1/4+5/4 ln|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 (x+2)/(2x-1)

\int \frac{x + 2}{2 x - 1} d x

\frac{1}{2} x - \frac{1}{4} + \frac{5}{4} ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{x + 2}{2 x - 1} d x

Expand

\frac{x + 2}{2 x - 1} : \frac{x}{2 x - 1} + \frac{2}{2 x - 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{2 x - 1} d x + \int \frac{2}{2 x - 1} d x

\int \frac{x}{2 x - 1} d x = \frac{1}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣)

\int \frac{2}{2 x - 1} d x = ln ∣ 2 x - 1 ∣

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

Simplify

\frac{1}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣) + ln ∣ 2 x - 1 ∣ : \frac{1}{2} x - \frac{1}{4} + \frac{5}{4} ln ∣ 2 x - 1 ∣

= \frac{1}{2} x - \frac{1}{4} + \frac{5}{4} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= \frac{1}{2} x - \frac{1}{4} + \frac{5}{4} ln ∣ 2 x - 1 ∣ + C

y = e^{3 x}, y = e^{4 x}, x = 1

x = 7 - 7 y^{2}, x = 7 y^{2} - 7

2 t y^{2} d t + 2 t^{2} y d y = 0

\int (e^{c o s (x)} sin (x)) d x

x \to 0 lim (ln (x^{2}) 2 e^{x} + 1)

What is the integral of (x+2)/(2x-1) ?
The integral of (x+2)/(2x-1) is 1/2 x-1/4+5/4 ln|2x-1|+C