What is the integral of (x^2+6x+6)/(x^3+9x^2+18x+7) ?

The integral of (x^2+6x+6)/(x^3+9x^2+18x+7) is 1/3 ln|x^3+9x^2+18x+7|+C

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integral of (x^2+6x+6)/(x^3+9x^2+18x+7)

\int \frac{x ^{2} + 6 x + 6}{x ^{3} + 9 x ^{2} + 18 x + 7} d x

\frac{1}{3} ln x^{3} + 9 x^{2} + 18 x + 7 + C

Solution steps

\int \frac{x ^{2} + 6 x + 6}{x ^{3} + 9 x ^{2} + 18 x + 7} 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} + 9 x^{2} + 18 x + 7

= \frac{1}{3} ln x^{3} + 9 x^{2} + 18 x + 7

Add a constant to the solution

= \frac{1}{3} ln x^{3} + 9 x^{2} + 18 x + 7 + C

\frac{d}{d x} (\frac{54}{( 3 x + 1 ) ^{4}})

x^{2} y^{^{'}} + 2 x y = 9 cos^{2} (x)

\frac{\partial}{\partial t} (x^{2} sin (3 t))

\frac{d}{d x} (\frac{( x ^{2} - 1 )}{2 x - 3})

x \to 1 + lim (∣ x ∣)

What is the integral of (x^2+6x+6)/(x^3+9x^2+18x+7) ?
The integral of (x^2+6x+6)/(x^3+9x^2+18x+7) is 1/3 ln|x^3+9x^2+18x+7|+C