What is the integral of (x^2+8x+1)/(x^3+12x^2+3x+6) ?

The integral of (x^2+8x+1)/(x^3+12x^2+3x+6) is 1/3 ln|x^3+12x^2+3x+6|+C

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integral of (x^2+8x+1)/(x^3+12x^2+3x+6)

\int \frac{x ^{2} + 8 x + 1}{x ^{3} + 12 x ^{2} + 3 x + 6} d x

\frac{1}{3} ln x^{3} + 12 x^{2} + 3 x + 6 + C

Solution steps

\int \frac{x ^{2} + 8 x + 1}{x ^{3} + 12 x ^{2} + 3 x + 6} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} + 12 x^{2} + 3 x + 6

= \frac{1}{3} ln x^{3} + 12 x^{2} + 3 x + 6

Add a constant to the solution

= \frac{1}{3} ln x^{3} + 12 x^{2} + 3 x + 6 + C

\int \frac{- 4 y - 1}{y + 1} d y

f (x) = x (3 x - 1)

\frac{d}{d x} ((3 x + 4)^{6})

\frac{d}{d x} (- e^{7 x})

\frac{d}{d x} (5 cos (π x))

What is the integral of (x^2+8x+1)/(x^3+12x^2+3x+6) ?
The integral of (x^2+8x+1)/(x^3+12x^2+3x+6) is 1/3 ln|x^3+12x^2+3x+6|+C