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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

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