What is the integral of 1/(7-4x-x^2) ?

The integral of 1/(7-4x-x^2) is 1/(2sqrt(11))(ln|(x+2)/(sqrt(11))+1|-ln|(x+2)/(sqrt(11))-1|)+C

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integral of 1/(7-4x-x^2)

\int \frac{1}{7 - 4 x - x ^{2}} d x

\frac{1}{2 11} (ln \frac{x + 2}{11} + 1 - ln \frac{x + 2}{11} - 1) + C

Solution steps

\int \frac{1}{7 - 4 x - x ^{2}} d x

Complete the square

7 - 4 x - x^{2} : - (x + 2)^{2} + 11

= \int \frac{1}{- ( x + 2 ) ^{2} + 11} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 11} d u

= \int \frac{1}{- u ^{2} + 11} d u

Apply Integral Substitution

= \int \frac{1}{11 ( - 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}{11} \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}{11} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{11} \frac{ln \frac{x + 2}{11} + 1}{2} - \frac{ln \frac{x + 2}{11} - 1}{2}

Simplify

\frac{1}{11} \frac{ln \frac{x + 2}{11} + 1}{2} - \frac{ln \frac{x + 2}{11} - 1}{2} : \frac{1}{2 11} (ln \frac{x + 2}{11} + 1 - ln \frac{x + 2}{11} - 1)

= \frac{1}{2 11} (ln \frac{x + 2}{11} + 1 - ln \frac{x + 2}{11} - 1)

Add a constant to the solution

= \frac{1}{2 11} (ln \frac{x + 2}{11} + 1 - ln \frac{x + 2}{11} - 1) + C

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

x = 3 y, x = 2 + y^{2}

6 u^{- 1}

x \to 0 lim (\frac{( e ^{x} )}{x})

38 x e^{x^{2}}

What is the integral of 1/(7-4x-x^2) ?
The integral of 1/(7-4x-x^2) is 1/(2sqrt(11))(ln|(x+2)/(sqrt(11))+1|-ln|(x+2)/(sqrt(11))-1|)+C