What is the integral of 1/(3-6x-x^2) ?

The integral of 1/(3-6x-x^2) is 1/(4sqrt(3))(ln|(x+3)/(2sqrt(3))+1|-ln|(x+3)/(2sqrt(3))-1|)+C

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integral of 1/(3-6x-x^2)

\int \frac{1}{3 - 6 x - x ^{2}} d x

\frac{1}{4 3} (ln \frac{x + 3}{2 3} + 1 - ln \frac{x + 3}{2 3} - 1) + C

Solution steps

\int \frac{1}{3 - 6 x - x ^{2}} d x

Complete the square

3 - 6 x - x^{2} : - (x + 3)^{2} + 12

= \int \frac{1}{- ( x + 3 ) ^{2} + 12} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 12} d u

= \int \frac{1}{- u ^{2} + 12} d u

Apply Integral Substitution

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

Substitute back

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

Simplify

\frac{1}{2 3} \frac{ln \frac{x + 3}{2 3} + 1}{2} - \frac{ln \frac{x + 3}{2 3} - 1}{2} : \frac{1}{4 3} (ln \frac{x + 3}{2 3} + 1 - ln \frac{x + 3}{2 3} - 1)

= \frac{1}{4 3} (ln \frac{x + 3}{2 3} + 1 - ln \frac{x + 3}{2 3} - 1)

Add a constant to the solution

= \frac{1}{4 3} (ln \frac{x + 3}{2 3} + 1 - ln \frac{x + 3}{2 3} - 1) + C

x \to 0 lim (5 x + 4)

\frac{\partial}{\partial x} (ln (5 x^{2} + 4 y^{2} + 7))

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

\int cos^{5} (x) sin (x) d x

z

What is the integral of 1/(3-6x-x^2) ?
The integral of 1/(3-6x-x^2) is 1/(4sqrt(3))(ln|(x+3)/(2sqrt(3))+1|-ln|(x+3)/(2sqrt(3))-1|)+C