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

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

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integral of (x^2+2x+3)/(x^3+3x^2+9x+1)

\int \frac{x ^{2} + 2 x + 3}{x ^{3} + 3 x ^{2} + 9 x + 1} d x

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

Solution steps

\int \frac{x ^{2} + 2 x + 3}{x ^{3} + 3 x ^{2} + 9 x + 1} 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 + 1

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

Add a constant to the solution

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

\int \frac{1}{2 x ^{2} + x} d x

sin (x), - sin (x), [0, 4]

\int \frac{10 x ^{3} - 5 x ^{2} + 4 x}{x ^{2}} d x

\frac{d}{d x} (tanh (2 x - 3))

f (x) = x^{2} + x, (1, 2)

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