What is the integral of (3x^2+1)/(x^3+x-1) ?

The integral of (3x^2+1)/(x^3+x-1) is ln|x^3+x-1|+C

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integral of (3x^2+1)/(x^3+x-1)

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

ln x^{3} + x - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{3} + x - 1

= ln x^{3} + x - 1

Add a constant to the solution

= ln x^{3} + x - 1 + C

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What is the integral of (3x^2+1)/(x^3+x-1) ?
The integral of (3x^2+1)/(x^3+x-1) is ln|x^3+x-1|+C