What is the integral of 1/(x^5+16x^3+64x) ?

The integral of 1/(x^5+16x^3+64x) is 1/64 ln|x|-1/128 ln|x^2+8|+1/(16(x^2+8))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^5+16x^3+64x)

\int \frac{1}{x ^{5} + 16 x ^{3} + 64 x} d x

\frac{1}{64} ln ∣ x ∣ - \frac{1}{128} ln x^{2} + 8 + \frac{1}{16 ( x ^{2} + 8 )} + C

Solution steps

\int \frac{1}{x ^{5} + 16 x ^{3} + 64 x} d x

Take the partial fraction of

\frac{1}{x ^{5} + 16 x ^{3} + 64 x} : \frac{1}{64 x} - \frac{x}{64 ( x ^{2} + 8 )} - \frac{x}{8 ( x ^{2} + 8 ) ^{2}}

= \int \frac{1}{64 x} - \frac{x}{64 ( x ^{2} + 8 )} - \frac{x}{8 ( x ^{2} + 8 ) ^{2}} 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}{64 x} d x - \int \frac{x}{64 ( x ^{2} + 8 )} d x - \int \frac{x}{8 ( x ^{2} + 8 ) ^{2}} d x

\int \frac{1}{64 x} d x = \frac{1}{64} ln ∣ x ∣

\int \frac{x}{64 ( x ^{2} + 8 )} d x = \frac{1}{128} ln x^{2} + 8

\int \frac{x}{8 ( x ^{2} + 8 ) ^{2}} d x = - \frac{1}{16 ( x ^{2} + 8 )}

= \frac{1}{64} ln ∣ x ∣ - \frac{1}{128} ln x^{2} + 8 - (- \frac{1}{16 ( x ^{2} + 8 )})

Simplify

= \frac{1}{64} ln ∣ x ∣ - \frac{1}{128} ln x^{2} + 8 + \frac{1}{16 ( x ^{2} + 8 )}

Add a constant to the solution

= \frac{1}{64} ln ∣ x ∣ - \frac{1}{128} ln x^{2} + 8 + \frac{1}{16 ( x ^{2} + 8 )} + C

3 x + 7, at x = 3

x \to 2 lim (6 x + 1)

\int_{- \infty}^{- 1} \frac{1}{x ^{2}} d x

\int \frac{x ^{2} + 1}{x} \cdot ln (x) d x

\frac{3}{2 x + 4} + 9 x^{2}

What is the integral of 1/(x^5+16x^3+64x) ?
The integral of 1/(x^5+16x^3+64x) is 1/64 ln|x|-1/128 ln|x^2+8|+1/(16(x^2+8))+C