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

The integral of x/(x^4+3x^2-4) is 1/10 ln|x+1|+1/10 ln|x-1|-1/10 ln|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/(x^4+3x^2-4)

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

\frac{1}{10} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln x^{2} + 4 + C

Solution steps

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

Take the partial fraction of

\frac{x}{x ^{4} + 3 x ^{2} - 4} : \frac{1}{10 ( x + 1 )} + \frac{1}{10 ( x - 1 )} - \frac{x}{5 ( x ^{2} + 4 )}

= \int \frac{1}{10 ( x + 1 )} + \frac{1}{10 ( x - 1 )} - \frac{x}{5 ( 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 \frac{1}{10 ( x + 1 )} d x + \int \frac{1}{10 ( x - 1 )} d x - \int \frac{x}{5 ( x ^{2} + 4 )} d x

\int \frac{1}{10 ( x + 1 )} d x = \frac{1}{10} ln ∣ x + 1 ∣

\int \frac{1}{10 ( x - 1 )} d x = \frac{1}{10} ln ∣ x - 1 ∣

\int \frac{x}{5 ( x ^{2} + 4 )} d x = \frac{1}{10} ln x^{2} + 4

= \frac{1}{10} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln x^{2} + 4

Add a constant to the solution

= \frac{1}{10} ln ∣ x + 1 ∣ + \frac{1}{10} ln ∣ x - 1 ∣ - \frac{1}{10} ln x^{2} + 4 + C

n = 1 \sum \infty \frac{n + 2}{n !}

\frac{d N}{d t} = N (1 - 0.0004 N)

\int_{0}^{1} \frac{x}{x ^{2} + x + 1} d x

\frac{d x}{d t} + 5 t^{7} x^{9} + \frac{x}{t} = 0

x \to 0 - lim (\frac{ln ( x ^{2} )}{x})

What is the integral of x/(x^4+3x^2-4) ?
The integral of x/(x^4+3x^2-4) is 1/10 ln|x+1|+1/10 ln|x-1|-1/10 ln|x^2+4|+C