What is the integral of (2x^2-3x-36)/((2x-1)(x^2+9)) ?

The integral of (2x^2-3x-36)/((2x-1)(x^2+9)) is -2ln|2x-1|+3/2 ln|x^2+9|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^2-3x-36)/((2x-1)(x^2+9))

\int \frac{2 x ^{2} - 3 x - 36}{( 2 x - 1 ) ( x ^{2} + 9 )} d x

- 2 ln ∣ 2 x - 1 ∣ + \frac{3}{2} ln x^{2} + 9 + C

Solution steps

\int \frac{2 x ^{2} - 3 x - 36}{( 2 x - 1 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{2 x ^{2} - 3 x - 36}{( 2 x - 1 ) ( x ^{2} + 9 )} : - \frac{4}{2 x - 1} + \frac{3 x}{x ^{2} + 9}

= \int - \frac{4}{2 x - 1} + \frac{3 x}{x ^{2} + 9} 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{4}{2 x - 1} d x + \int \frac{3 x}{x ^{2} + 9} d x

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

\int \frac{3 x}{x ^{2} + 9} d x = \frac{3}{2} ln x^{2} + 9

= - 2 ln ∣ 2 x - 1 ∣ + \frac{3}{2} ln x^{2} + 9

Add a constant to the solution

= - 2 ln ∣ 2 x - 1 ∣ + \frac{3}{2} ln x^{2} + 9 + C

\frac{d}{d x} (1 - ln (x))

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

x \to 3 lim (\frac{x + 1}{x - 2})

g (x) = 15 x

\frac{d}{d x} (14)

What is the integral of (2x^2-3x-36)/((2x-1)(x^2+9)) ?
The integral of (2x^2-3x-36)/((2x-1)(x^2+9)) is -2ln|2x-1|+3/2 ln|x^2+9|+C