What is the integral of (4x^3-1)/(x^4-x+4) ?

The integral of (4x^3-1)/(x^4-x+4) is ln|x^4-x+4|+C

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integral of (4x^3-1)/(x^4-x+4)

\int \frac{4 x ^{3} - 1}{x ^{4} - x + 4} d x

ln x^{4} - x + 4 + C

Solution steps

\int \frac{4 x ^{3} - 1}{x ^{4} - x + 4} d x

Apply u-substitution

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{4} - x + 4

= ln x^{4} - x + 4

Add a constant to the solution

= ln x^{4} - x + 4 + C

What is the integral of (4x^3-1)/(x^4-x+4) ?
The integral of (4x^3-1)/(x^4-x+4) is ln|x^4-x+4|+C