What is the integral of 1/(8-2x-x^2) ?

The integral of 1/(8-2x-x^2) is 1/6 (ln|(x+1)/3+1|-ln|(x+1)/3-1|)+C

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integral of 1/(8-2x-x^2)

\int \frac{1}{8 - 2 x - x ^{2}} d x

\frac{1}{6} (ln \frac{x + 1}{3} + 1 - ln \frac{x + 1}{3} - 1) + C

Solution steps

\int \frac{1}{8 - 2 x - x ^{2}} d x

Complete the square

8 - 2 x - x^{2} : - (x + 1)^{2} + 9

= \int \frac{1}{- ( x + 1 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 9} d u

= \int \frac{1}{- u ^{2} + 9} 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

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

Simplify

\frac{1}{3} (\frac{ln \frac{x + 1}{3} + 1}{2} - \frac{ln \frac{x + 1}{3} - 1}{2}) : \frac{1}{6} (ln \frac{x + 1}{3} + 1 - ln \frac{x + 1}{3} - 1)

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

Add a constant to the solution

= \frac{1}{6} (ln \frac{x + 1}{3} + 1 - ln \frac{x + 1}{3} - 1) + C

\frac{d}{d x} (e^{x^{2} + 2 x + 1})

x \to 4 lim (x + 5)

f (x) = - 2 x^{2}

(y^{2} + y x) d x + (- x^{2}) d y = 0

y^{^{'}} = \frac{y ^{2} + 4 x x ^{2} + y ^{2}}{x y}

What is the integral of 1/(8-2x-x^2) ?
The integral of 1/(8-2x-x^2) is 1/6 (ln|(x+1)/3+1|-ln|(x+1)/3-1|)+C