What is the integral of 1/(x(2+x^8)) ?

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

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integral of 1/(x(2+x^8))

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

\frac{1}{8} (4 ln (x) - \frac{1}{2} ln x^{8} + 2) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{8 u ( 2 + u )} d u

= \int \frac{1}{8 u ( 2 + u )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u ( 2 + u )} d u

Take the partial fraction of

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

= \frac{1}{8} \cdot \int \frac{1}{2 u} - \frac{1}{2 ( u + 2 )} 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}{8} (\int \frac{1}{2 u} d u - \int \frac{1}{2 ( u + 2 )} d u)

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

\int \frac{1}{2 ( u + 2 )} d u = \frac{1}{2} ln ∣ u + 2 ∣

= \frac{1}{8} (\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣)

Substitute back

u = x^{8}

= \frac{1}{8} (\frac{1}{2} ln x^{8} - \frac{1}{2} ln x^{8} + 2)

\frac{1}{2} ln x^{8} = 4 ln (x)

= \frac{1}{8} (4 ln (x) - \frac{1}{2} ln x^{8} + 2)

Add a constant to the solution

= \frac{1}{8} (4 ln (x) - \frac{1}{2} ln x^{8} + 2) + C

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