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

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

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integral of (x^{n-1})/(1+x^n)

\int \frac{x ^{n - 1}}{1 + x ^{n}} d x

\frac{1}{n} ln ∣ 1 + x^{n} ∣ + C

Solution steps

\int \frac{x ^{n - 1}}{1 + x ^{n}} d x

Apply u-substitution

= \int \frac{1}{n u} 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} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{n} ln ∣ u ∣

Substitute back

u = 1 + x^{n}

= \frac{1}{n} ln ∣ 1 + x^{n} ∣

Add a constant to the solution

= \frac{1}{n} ln ∣ 1 + x^{n} ∣ + C

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