What is the integral of (3x^2)/(x^4-1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x^2)/(x^4-1)

\int \frac{3 x ^{2}}{x ^{4} - 1} d x

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

Solution steps

\int \frac{3 x ^{2}}{x ^{4} - 1} d x

Take the constant out:

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

= 3 \cdot \int \frac{x ^{2}}{x ^{4} - 1} d x

Take the partial fraction of

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

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

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

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

= 3 (\frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣)

Add a constant to the solution

= 3 (\frac{1}{2} arctan (x) - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) + C

2 x^{2} - 7

\frac{d}{d y} (\frac{A}{y ^{11}} + B e^{y})

\int \frac{2 x ^{3} - 5 x ^{2} + x - 6}{x} d x

x \to 0 lim (\frac{16}{x ^{4}})

f (x) = \frac{1}{x + 6}

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