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

The integral of (4x^2-1)/((x^2+4)(2-x)) is -17/8 (1/2 ln|(x^2)/4+1|+arctan(x/2))-15/8 ln|x-2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4x^2-1)/((x^2+4)(2-x))

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

- \frac{17}{8} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 + arctan (\frac{x}{2})) - \frac{15}{8} ln ∣ x - 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{4 x ^{2} - 1}{( x ^{2} + 4 ) ( 2 - x )} : \frac{- 17 x - 34}{8 ( x ^{2} + 4 )} - \frac{15}{8 ( x - 2 )}

= \int \frac{- 17 x - 34}{8 ( x ^{2} + 4 )} - \frac{15}{8 ( x - 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{- 17 x - 34}{8 ( x ^{2} + 4 )} d x - \int \frac{15}{8 ( x - 2 )} d x

\int \frac{- 17 x - 34}{8 ( x ^{2} + 4 )} d x = - \frac{17}{8} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 + arctan (\frac{x}{2}))

\int \frac{15}{8 ( x - 2 )} d x = \frac{15}{8} ln ∣ x - 2 ∣

= - \frac{17}{8} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 + arctan (\frac{x}{2})) - \frac{15}{8} ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{17}{8} (\frac{1}{2} ln \frac{x ^{2}}{4} + 1 + arctan (\frac{x}{2})) - \frac{15}{8} ln ∣ x - 2 ∣ + C

f (x) = - \frac{x}{4}

f (x) = 4 x^{2} - 4 x + 1, at x = 1, y = 0

\int cos (4 x) sin (x) d x

y = \frac{x ^{2} + 1}{x ^{3} - 1} - 5 x^{2}

\frac{e ^{x}}{1 + e ^{x}}

What is the integral of (4x^2-1)/((x^2+4)(2-x)) ?
The integral of (4x^2-1)/((x^2+4)(2-x)) is -17/8 (1/2 ln|(x^2)/4+1|+arctan(x/2))-15/8 ln|x-2|+C