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

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

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

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

\frac{1}{3} ln x^{3} + 9 x^{2} + 6 x + 9 + C

Solution steps

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

= \frac{1}{3} ln x^{3} + 9 x^{2} + 6 x + 9

Add a constant to the solution

= \frac{1}{3} ln x^{3} + 9 x^{2} + 6 x + 9 + C

y^{^{'}} + 0.06 y = 6

n = 1 \sum \infty \frac{( n ! )}{n ^{n}}

y = 91

\int x^{2} (x + 1) d x

x \to - 3 lim (\frac{x + 3}{x + 4})

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