What is the integral of 1/(-x^2-10x-21) ?

The integral of 1/(-x^2-10x-21) is 1/4 (ln|(x+5)/2+1|-ln|(x+5)/2-1|)+C

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integral of 1/(-x^2-10x-21)

\int \frac{1}{- x ^{2} - 10 x - 21} d x

\frac{1}{4} (ln \frac{x + 5}{2} + 1 - ln \frac{x + 5}{2} - 1) + C

Solution steps

\int \frac{1}{- x ^{2} - 10 x - 21} d x

Complete the square

- x^{2} - 10 x - 21 : - (x + 5)^{2} + 4

= \int \frac{1}{- ( x + 5 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 4} d u

= \int \frac{1}{- u ^{2} + 4} d u

Apply Integral Substitution

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

Substitute back

= \frac{1}{2} (\frac{ln \frac{x + 5}{2} + 1}{2} - \frac{ln \frac{x + 5}{2} - 1}{2})

Simplify

\frac{1}{2} (\frac{ln \frac{x + 5}{2} + 1}{2} - \frac{ln \frac{x + 5}{2} - 1}{2}) : \frac{1}{4} (ln \frac{x + 5}{2} + 1 - ln \frac{x + 5}{2} - 1)

= \frac{1}{4} (ln \frac{x + 5}{2} + 1 - ln \frac{x + 5}{2} - 1)

Add a constant to the solution

= \frac{1}{4} (ln \frac{x + 5}{2} + 1 - ln \frac{x + 5}{2} - 1) + C

(x + 4) (x - 5)^{2}

\frac{s + 1}{( s + 2 ) ( s + 6 ) ^{2}}

f (x) = x^{3} - 2 x + 7, at x = - 2

x^{4} - 2 x^{2} + 5

n = 1 \sum \infty 8 e^{- 6 n}

What is the integral of 1/(-x^2-10x-21) ?
The integral of 1/(-x^2-10x-21) is 1/4 (ln|(x+5)/2+1|-ln|(x+5)/2-1|)+C