What is the integral of (3x^2-8x+1)/(x^3-4x^2+x+6) ?

The integral of (3x^2-8x+1)/(x^3-4x^2+x+6) is ln|x^3-4x^2+x+6|+C

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integral of (3x^2-8x+1)/(x^3-4x^2+x+6)

\int \frac{3 x ^{2} - 8 x + 1}{x ^{3} - 4 x ^{2} + x + 6} d x

ln x^{3} - 4 x^{2} + x + 6 + C

Solution steps

\int \frac{3 x ^{2} - 8 x + 1}{x ^{3} - 4 x ^{2} + x + 6} 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} - 4 x^{2} + x + 6

= ln x^{3} - 4 x^{2} + x + 6

Add a constant to the solution

= ln x^{3} - 4 x^{2} + x + 6 + C

\int sin (2 x) sin (4 x) sin (6 x) d x

\int \frac{x ^{\frac{3}{2}} - 2 x ^{6} - 1.2}{x} d x

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

x \to 0 - lim (\frac{14 x ^{4} - 14 x}{x})

\frac{\partial}{\partial x} (x^{3} y + sin (x y))

What is the integral of (3x^2-8x+1)/(x^3-4x^2+x+6) ?
The integral of (3x^2-8x+1)/(x^3-4x^2+x+6) is ln|x^3-4x^2+x+6|+C