What is the integral of 1/(x^5+2x^3+x) ?

The integral of 1/(x^5+2x^3+x) is ln|x|-1/2 ln|x^2+1|+1/(2(x^2+1))+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+2x^3+x)

\int \frac{1}{x ^{5} + 2 x ^{3} + x} d x

ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1 + \frac{1}{2 ( x ^{2} + 1 )} + C

Solution steps

\int \frac{1}{x ^{5} + 2 x ^{3} + x} d x

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

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

= ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1 - (- \frac{1}{2 ( x ^{2} + 1 )})

Simplify

= ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1 + \frac{1}{2 ( x ^{2} + 1 )}

Add a constant to the solution

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

\int_{0}^{10} 400 - 20 x d x

f (t) = 5 t

\int 4 - y d y

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

\frac{d}{d x} (\frac{4}{5} x^{\frac{5}{4}})

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