What is the integral of 1/(x(x^7+8)) ?

The integral of 1/(x(x^7+8)) is 1/7 (1/8 ln|x^7|-1/8 ln|x^7+8|)+C

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integral of 1/(x(x^7+8))

\int \frac{1}{x ( x ^{7} + 8 )} d x

\frac{1}{7} (\frac{1}{8} ln x^{7} - \frac{1}{8} ln x^{7} + 8) + C

Solution steps

\int \frac{1}{x ( x ^{7} + 8 )} d x

Apply u-substitution

= \int \frac{1}{7 u ( u + 8 )} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{1}{u ( u + 8 )} d u

Take the partial fraction of

\frac{1}{u ( u + 8 )} : \frac{1}{8 u} - \frac{1}{8 ( u + 8 )}

= \frac{1}{7} \cdot \int \frac{1}{8 u} - \frac{1}{8 ( u + 8 )} 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}{7} (\int \frac{1}{8 u} d u - \int \frac{1}{8 ( u + 8 )} d u)

\int \frac{1}{8 u} d u = \frac{1}{8} ln ∣ u ∣

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

= \frac{1}{7} (\frac{1}{8} ln ∣ u ∣ - \frac{1}{8} ln ∣ u + 8 ∣)

Substitute back

u = x^{7}

= \frac{1}{7} (\frac{1}{8} ln x^{7} - \frac{1}{8} ln x^{7} + 8)

Add a constant to the solution

= \frac{1}{7} (\frac{1}{8} ln x^{7} - \frac{1}{8} ln x^{7} + 8) + C

What is the integral of 1/(x(x^7+8)) ?
The integral of 1/(x(x^7+8)) is 1/7 (1/8 ln|x^7|-1/8 ln|x^7+8|)+C