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

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

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integral of x/(x^8-1)

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

Apply u-substitution

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

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

Take the partial fraction of

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

y^{^{''}} + 9 y^{^{'}} + 18 y = 2050 e^{2 t} cos (10 t)

\int_{- 3}^{1} x + 3 d x

y = 2 x^{3} - 3 x + 1, (1, 0)

\frac{d}{d x} (- x sin (x) + cos (x))

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