What is the integral of x/(1-x^8) ?

The integral of x/(1-x^8) is 1/2 (1/2 arctan(x^2)+1/4 ln|x^2+1|-1/4 ln|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 x/(1-x^8)

\int \frac{x}{1 - x ^{8}} d x

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

Solution steps

\int \frac{x}{1 - x ^{8}} d x

Apply u-substitution

: \int \frac{1}{2 ( 1 - u ^{4} )} d u

= \int \frac{1}{2 ( 1 - u ^{4} )} d u

Take the constant out:

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

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

Take the partial fraction of

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

= \frac{1}{2} \cdot \int \frac{1}{2 ( u ^{2} + 1 )} + \frac{1}{4 ( u + 1 )} - \frac{1}{4 ( u - 1 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

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

Substitute back

u = x^{2}

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

Add a constant to the solution

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

f (x) = cos (x^{2}) 2 x

\int 8 x^{- 1} d x

\int \frac{1}{x ^{2} + 8} d x

f (x) = - 2 x^{3}

\frac{s - 1}{2 s ^{2} + s + 6}

What is the integral of x/(1-x^8) ?
The integral of x/(1-x^8) is 1/2 (1/2 arctan(x^2)+1/4 ln|x^2+1|-1/4 ln|x^2-1|)+C