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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

\int \frac{x ^{n}}{n} d x

\int_{5}^{9} (\frac{x - 4}{x ^{2}}) d x

\frac{d}{d x} (\frac{x ^{2} + 4 x + 5 x ^{3}}{x + 1})

\int_{1}^{2} 2^{- x} d x

\int_{0}^{1} \frac{25}{4 y - 1} d y

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