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

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

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

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

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

Solution steps

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

Apply u-substitution

= \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 + 3

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

Add a constant to the solution

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

t an g e n t - 3 cos (x), at x = - \frac{π}{3}

ma c l a u r in e^{2 x^{3}}

\frac{\partial}{\partial y} (4 x^{3} y^{2} - 1)

l a pl a ce t r an s f or m cos^{2} (13 t)

d er i v a t i v e y = ln (5)

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