What is the integral of 9/(x^4-81) ?

The integral of 9/(x^4-81) is 9(-1/54 arctan(x/3)-1/108 ln|x+3|+1/108 ln|x-3|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9/(x^4-81)

\int \frac{9}{x ^{4} - 81} d x

9 (- \frac{1}{54} arctan (\frac{x}{3}) - \frac{1}{108} ln ∣ x + 3 ∣ + \frac{1}{108} ln ∣ x - 3 ∣) + C

Solution steps

\int \frac{9}{x ^{4} - 81} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{x ^{4} - 81} d x

Take the partial fraction of

\frac{1}{x ^{4} - 81} : - \frac{1}{18 ( x ^{2} + 9 )} - \frac{1}{108 ( x + 3 )} + \frac{1}{108 ( x - 3 )}

= 9 \cdot \int - \frac{1}{18 ( x ^{2} + 9 )} - \frac{1}{108 ( x + 3 )} + \frac{1}{108 ( x - 3 )} 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

= 9 (- \int \frac{1}{18 ( x ^{2} + 9 )} d x - \int \frac{1}{108 ( x + 3 )} d x + \int \frac{1}{108 ( x - 3 )} d x)

\int \frac{1}{18 ( x ^{2} + 9 )} d x = \frac{1}{54} arctan (\frac{x}{3})

\int \frac{1}{108 ( x + 3 )} d x = \frac{1}{108} ln ∣ x + 3 ∣

\int \frac{1}{108 ( x - 3 )} d x = \frac{1}{108} ln ∣ x - 3 ∣

= 9 (- \frac{1}{54} arctan (\frac{x}{3}) - \frac{1}{108} ln ∣ x + 3 ∣ + \frac{1}{108} ln ∣ x - 3 ∣)

Add a constant to the solution

= 9 (- \frac{1}{54} arctan (\frac{x}{3}) - \frac{1}{108} ln ∣ x + 3 ∣ + \frac{1}{108} ln ∣ x - 3 ∣) + C

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

\int e^{9 x - 8} d x

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

\int 3 x + 8 d x

\int e^{- x} (x + 1) d x

What is the integral of 9/(x^4-81) ?
The integral of 9/(x^4-81) is 9(-1/54 arctan(x/3)-1/108 ln|x+3|+1/108 ln|x-3|)+C