What is the integral of (x^3)/((x^2-4)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3)/((x^2-4))

integral

\frac{x ^{3}}{( x ^{2} - 4 )}

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

Solution steps

\int \frac{x ^{3}}{x ^{2} - 4} d x

Apply u-substitution

: \int \frac{u + 4}{2 u} d u

= \int \frac{u + 4}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u + 4}{u} d u

Expand

\frac{u + 4}{u} : 1 + \frac{4}{u}

= \frac{1}{2} \cdot \int 1 + \frac{4}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d u + \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= \frac{1}{2} (u + 4 ln ∣ u ∣)

Substitute back

u = x^{2} - 4

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

Add a constant to the solution

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

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

\frac{x}{( x ^{2} - x + 10 )}

\frac{1}{ln ( 1 + 3 x )}

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

\frac{( x - 1 )}{( x + 1 )}

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