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

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

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

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

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

Solution steps

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

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

Simplify

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

Add a constant to the solution

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

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