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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{1}{x} e^{x^{2}} d x

\int \frac{- 1}{1 + e ^{- x}} d x

\int 5 x^{2} \cdot x d x

\int \frac{3 x ^{2}}{2 ( 9 x ^{2} + 1 )} d x

\int \frac{e ^{2 t}}{t} d t

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