What is the integral of 1/(x^2+6x+6) ?

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

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integral of 1/(x^2+6x+6)

\int \frac{1}{x ^{2} + 6 x + 6} d x

- \frac{1}{2 3} (ln \frac{x + 3}{3} + 1 - ln \frac{x + 3}{3} - 1) + C

Solution steps

\int \frac{1}{x ^{2} + 6 x + 6} d x

Complete the square

x^{2} + 6 x + 6 : (x + 3)^{2} - 3

= \int \frac{1}{( x + 3 ) ^{2} - 3} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 3} d u

= \int \frac{1}{u ^{2} - 3} d u

Apply Integral Substitution

= \int \frac{1}{3 ( v ^{2} - 1 )} d v

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}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= \frac{1}{3} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{3} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{3} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= \frac{1}{3} - \frac{ln \frac{x + 3}{3} + 1}{2} - \frac{ln \frac{x + 3}{3} - 1}{2}

Simplify

\frac{1}{3} - \frac{ln \frac{x + 3}{3} + 1}{2} - \frac{ln \frac{x + 3}{3} - 1}{2} : - \frac{1}{2 3} (ln \frac{x + 3}{3} + 1 - ln \frac{x + 3}{3} - 1)

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

Add a constant to the solution

= - \frac{1}{2 3} (ln \frac{x + 3}{3} + 1 - ln \frac{x + 3}{3} - 1) + C

\frac{d y}{d x} + 2 x y = x^{3}

\frac{\partial}{\partial y} (\frac{- y}{x ^{2} + 9 y ^{2}})

\int_{- 2}^{3} \frac{40}{x ^{4}} d x

y = x + 3

\frac{1}{s} - \frac{2}{( s + 3 ) ^{2} - 9}

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