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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Take the partial fraction of

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

Substitute back

u = 1 + x^{n}

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

Simplify

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

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

Add a constant to the solution

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

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