What is the integral of (x^2+10x+2)/(x^3+15x^2+6x+9) ?

The integral of (x^2+10x+2)/(x^3+15x^2+6x+9) is 1/3 ln|x^3+15x^2+6x+9|+C

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integral of (x^2+10x+2)/(x^3+15x^2+6x+9)

\int \frac{x ^{2} + 10 x + 2}{x ^{3} + 15 x ^{2} + 6 x + 9} d x

\frac{1}{3} ln x^{3} + 15 x^{2} + 6 x + 9 + C

Solution steps

\int \frac{x ^{2} + 10 x + 2}{x ^{3} + 15 x ^{2} + 6 x + 9} d x

Factor

x^{2} + 10 x + 2 : - (- x^{2} - 10 x - 2)

= \int \frac{- ( - x ^{2} - 10 x - 2 )}{x ^{3} + 15 x ^{2} + 6 x + 9} d x

Take the constant out:

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

= - \int \frac{- x ^{2} - 10 x - 2}{x ^{3} + 15 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} + 15 x^{2} + 6 x + 9

= - (- \frac{1}{3} ln x^{3} + 15 x^{2} + 6 x + 9)

Simplify

= \frac{1}{3} ln x^{3} + 15 x^{2} + 6 x + 9

Add a constant to the solution

= \frac{1}{3} ln x^{3} + 15 x^{2} + 6 x + 9 + C

f (x) = 3 - 2 x

x \to 0 lim (tan (2 x) csc (4 x))

y = x^{4} + 5 e^{x}, (0, 5)

\frac{d y}{d x} = π y + 12 e

x \to π lim (\frac{5 x}{3})

What is the integral of (x^2+10x+2)/(x^3+15x^2+6x+9) ?
The integral of (x^2+10x+2)/(x^3+15x^2+6x+9) is 1/3 ln|x^3+15x^2+6x+9|+C