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

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

integral

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

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

Solution steps

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

Expand

\frac{x ^{3} + 2}{4 - x ^{2}} : \frac{x ^{3}}{4 - x ^{2}} + \frac{2}{4 - x ^{2}}

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

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

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

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

Add a constant to the solution

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

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

16 - x^{2}

x^{2} \cdot (e^{x})^{2}

e^{x} - 2^{x}

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

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