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

The integral of (x+1)/(4x^2-4x+1) is 1/4 ln|2x-1|-3/(4(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)/(4x^2-4x+1)

\int \frac{x + 1}{4 x ^{2} - 4 x + 1} d x

\frac{1}{4} ln ∣ 2 x - 1 ∣ - \frac{3}{4 ( 2 x - 1 )} + C

Solution steps

\int \frac{x + 1}{4 x ^{2} - 4 x + 1} d x

Take the partial fraction of

\frac{x + 1}{4 x ^{2} - 4 x + 1} : \frac{1}{2 ( 2 x - 1 )} + \frac{3}{2 ( 2 x - 1 ) ^{2}}

= \int \frac{1}{2 ( 2 x - 1 )} + \frac{3}{2 ( 2 x - 1 ) ^{2}} 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

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

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

\int \frac{3}{2 ( 2 x - 1 ) ^{2}} d x = - \frac{3}{4 ( 2 x - 1 )}

= \frac{1}{4} ln ∣ 2 x - 1 ∣ - \frac{3}{4 ( 2 x - 1 )}

Add a constant to the solution

= \frac{1}{4} ln ∣ 2 x - 1 ∣ - \frac{3}{4 ( 2 x - 1 )} + C

\int \frac{1}{x} + 5 d x

\int \frac{x ^{\frac{1}{2}}}{8 ( 3 + x ^{\frac{3}{4}} )} d x

\int \frac{x ^{4} + 1}{x} d x

\int x^{7} e^{2 x} d x

\int \frac{1}{x - b} d x

What is the integral of (x+1)/(4x^2-4x+1) ?
The integral of (x+1)/(4x^2-4x+1) is 1/4 ln|2x-1|-3/(4(2x-1))+C