What is the integral of (5x-2)/((x-1)^2(2x+1)) ?

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

\int \frac{5 x - 2}{( x - 1 ) ^{2} ( 2 x + 1 )} d x

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

Solution steps

\int \frac{5 x - 2}{( x - 1 ) ^{2} ( 2 x + 1 )} d x

Take the partial fraction of

\frac{5 x - 2}{( x - 1 ) ^{2} ( 2 x + 1 )} : \frac{1}{x - 1} + \frac{1}{( x - 1 ) ^{2}} - \frac{2}{2 x + 1}

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

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

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

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

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

Add a constant to the solution

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

\int \frac{1}{4 x ln ( x ) - 3 x} d x

\int \frac{1}{2} t d t

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

\int \frac{12}{x ^{2} - 4} d x

\int \frac{15}{x ^{3} + x ^{2}} d x

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