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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

u = 1 + x^{8}

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

Simplify

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

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

Add a constant to the solution

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

\int e^{x} 1 - cos^{2} (x) d x

\int e^{- \frac{3 x}{4}} d x

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

\frac{2}{x ^{3}}

\int (1 - ∣ x ∣) d x

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