What is the integral of (x^5-x)/(x^8+1) ?

The integral of (x^5-x)/(x^8+1) is 1/(4sqrt(2))(-ln(2|x^4+sqrt(2)x^2+1|)+ln(2|x^4-sqrt(2)x^2+1|))+C

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integral of (x^5-x)/(x^8+1)

\int \frac{x ^{5} - x}{x ^{8} + 1} d x

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

Solution steps

\int \frac{x ^{5} - x}{x ^{8} + 1} d x

Apply u-substitution

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

Expand

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

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

\int \frac{u ^{2}}{u ^{4} + 1} d u = - \frac{1}{4 2} (ln 2 u^{2} + 22 u + 2 - 2 arctan (2 u + 1)) + \frac{1}{4 2} (ln 2 u^{2} - 22 u + 2 + 2 arctan (2 u - 1))

\int \frac{1}{u ^{4} + 1} d u = \frac{1}{4 2} (ln 2 u^{2} + 22 u + 2 + 2 arctan (2 u + 1)) + \frac{1}{4 2} (- ln 2 u^{2} - 22 u + 2 + 2 arctan (2 u - 1))

= \frac{1}{2} (- \frac{1}{4 2} (ln 2 u^{2} + 22 u + 2 - 2 arctan (2 u + 1)) + \frac{1}{4 2} (ln 2 u^{2} - 22 u + 2 + 2 arctan (2 u - 1)) - (\frac{1}{4 2} (ln 2 u^{2} + 22 u + 2 + 2 arctan (2 u + 1)) + \frac{1}{4 2} (- ln 2 u^{2} - 22 u + 2 + 2 arctan (2 u - 1))))

Substitute back

u = x^{2}

= \frac{1}{2} (- \frac{1}{4 2} (ln 2 (x^{2})^{2} + 22 x^{2} + 2 - 2 arctan (2 x^{2} + 1)) + \frac{1}{4 2} (ln 2 (x^{2})^{2} - 22 x^{2} + 2 + 2 arctan (2 x^{2} - 1)) - (\frac{1}{4 2} (ln 2 (x^{2})^{2} + 22 x^{2} + 2 + 2 arctan (2 x^{2} + 1)) + \frac{1}{4 2} (- ln 2 (x^{2})^{2} - 22 x^{2} + 2 + 2 arctan (2 x^{2} - 1))))

Simplify

\frac{1}{2} (- \frac{1}{4 2} (ln 2 (x^{2})^{2} + 22 x^{2} + 2 - 2 arctan (2 x^{2} + 1)) + \frac{1}{4 2} (ln 2 (x^{2})^{2} - 22 x^{2} + 2 + 2 arctan (2 x^{2} - 1)) - (\frac{1}{4 2} (ln 2 (x^{2})^{2} + 22 x^{2} + 2 + 2 arctan (2 x^{2} + 1)) + \frac{1}{4 2} (- ln 2 (x^{2})^{2} - 22 x^{2} + 2 + 2 arctan (2 x^{2} - 1)))) : \frac{1}{4 2} (- ln (2 x^{4} + 2 x^{2} + 1) + ln (2 x^{4} - 2 x^{2} + 1))

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

Add a constant to the solution

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

What is the integral of (x^5-x)/(x^8+1) ?
The integral of (x^5-x)/(x^8+1) is 1/(4sqrt(2))(-ln(2|x^4+sqrt(2)x^2+1|)+ln(2|x^4-sqrt(2)x^2+1|))+C