What is the integral of (x^4)/(x^4+5x^2+4) ?

The integral of (x^4)/(x^4+5x^2+4) is x+1/3 arctan(x)-8/3 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 (x^4)/(x^4+5x^2+4)

\int \frac{x ^{4}}{x ^{4} + 5 x ^{2} + 4} d x

x + \frac{1}{3} arctan (x) - \frac{8}{3} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{x ^{4}}{x ^{4} + 5 x ^{2} + 4} d x

Take the partial fraction of

\frac{x ^{4}}{x ^{4} + 5 x ^{2} + 4} : 1 + \frac{1}{3 ( x ^{2} + 1 )} - \frac{16}{3 ( x ^{2} + 4 )}

= \int 1 + \frac{1}{3 ( x ^{2} + 1 )} - \frac{16}{3 ( 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

= \int 1 d x + \int \frac{1}{3 ( x ^{2} + 1 )} d x - \int \frac{16}{3 ( x ^{2} + 4 )} d x

\int 1 d x = x

\int \frac{1}{3 ( x ^{2} + 1 )} d x = \frac{1}{3} arctan (x)

\int \frac{16}{3 ( x ^{2} + 4 )} d x = \frac{8}{3} arctan (\frac{x}{2})

= x + \frac{1}{3} arctan (x) - \frac{8}{3} arctan (\frac{x}{2})

Add a constant to the solution

= x + \frac{1}{3} arctan (x) - \frac{8}{3} arctan (\frac{x}{2}) + C

\int cos (x - y) d x

\int t cos (π t) d t

\int 1 + 3 x^{4} (12 x^{3}) d x

\int 2 x^{3} ln (x) d x

\int 72 (1 + x^{2}) (7 + 18 x + 6 x^{3})^{16} d x

What is the integral of (x^4)/(x^4+5x^2+4) ?
The integral of (x^4)/(x^4+5x^2+4) is x+1/3 arctan(x)-8/3 arctan(x/2)+C