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

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

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

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

\frac{1}{4} ln x^{4} - 4 x + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} - 4 x + 1

= \frac{1}{4} ln x^{4} - 4 x + 1

Add a constant to the solution

= \frac{1}{4} ln x^{4} - 4 x + 1 + C

\frac{d y}{d x} + (\frac{3}{x}) y = - 2 x - 7

3 sin (x), 4 cos (x), [0, 0.8 π]

\frac{d}{d x} (- 4 sec^{2} (x) + 6 sec^{4} (x))

x \to 3 lim (- 3 x - 7)

e^{u}

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