What is the integral of 2/(sqrt(-x^2-20x-9)) ?

The integral of 2/(sqrt(-x^2-20x-9)) is 2arcsin(1/(sqrt(91))(x+10))+C

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integral of 2/(sqrt(-x^2-20x-9))

\int \frac{2}{- x ^{2} - 20 x - 9} d x

2 arcsin (\frac{1}{91} (x + 10)) + C

Solution steps

\int \frac{2}{- x ^{2} - 20 x - 9} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{- x ^{2} - 20 x - 9} d x

Complete the square

- x^{2} - 20 x - 9 : - (x + 10)^{2} + 91

= 2 \cdot \int \frac{1}{- ( x + 10 ) ^{2} + 91} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 91} d u

= 2 \cdot \int \frac{1}{- u ^{2} + 91} d u

Apply Trigonometric Substitution

: \int 1 d v

= 2 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot 1 \cdot v

Substitute back

= 2 \cdot 1 \cdot arcsin (\frac{1}{91} (x + 10))

Simplify

= 2 arcsin (\frac{1}{91} (x + 10))

Add a constant to the solution

= 2 arcsin (\frac{1}{91} (x + 10)) + C

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What is the integral of 2/(sqrt(-x^2-20x-9)) ?
The integral of 2/(sqrt(-x^2-20x-9)) is 2arcsin(1/(sqrt(91))(x+10))+C