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

The integral of (x^2+x)/(2x-1) is 1/4 x^2+3/4 x+3/8 ln|2x-1|-1/4-3/16+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+x)/(2x-1)

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

\frac{1}{4} x^{2} + \frac{3}{4} x + \frac{3}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{4} - \frac{3}{16} + C

Solution steps

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

Expand

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

\int \frac{x ^{2}}{2 x - 1} d x = \frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ 2 x - 1 ∣ - 3)

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

= \frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ 2 x - 1 ∣ - 3) + \frac{1}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣)

Simplify

\frac{1}{16} (4 x^{2} + 4 x + 2 ln ∣ 2 x - 1 ∣ - 3) + \frac{1}{4} (2 x - 1 + ln ∣ 2 x - 1 ∣) : \frac{1}{4} x^{2} + \frac{3}{4} x + \frac{3}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{4} - \frac{3}{16}

= \frac{1}{4} x^{2} + \frac{3}{4} x + \frac{3}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{4} - \frac{3}{16}

Add a constant to the solution

= \frac{1}{4} x^{2} + \frac{3}{4} x + \frac{3}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{4} - \frac{3}{16} + C

\int (y + y^{2} cos (x)) d x

\int x^{\frac{1}{2}} + 7 d x

\int y (y + 1)^{2} d y

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

\int \frac{8}{x + 3 x} d x

What is the integral of (x^2+x)/(2x-1) ?
The integral of (x^2+x)/(2x-1) is 1/4 x^2+3/4 x+3/8 ln|2x-1|-1/4-3/16+C