What is the integral of 4/(sqrt(-4x^2-20x-9)) ?

The integral of 4/(sqrt(-4x^2-20x-9)) is 2arcsin(1/2 (x+5/2))+C

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integral of 4/(sqrt(-4x^2-20x-9))

\int \frac{4}{- 4 x ^{2} - 20 x - 9} d x

2 arcsin (\frac{1}{2} (x + \frac{5}{2})) + C

Solution steps

\int \frac{4}{- 4 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

= 4 \cdot \int \frac{1}{- 4 x ^{2} - 20 x - 9} d x

Complete the square

- 4 x^{2} - 20 x - 9 : - 4 (x + \frac{5}{2})^{2} + 16

= 4 \cdot \int \frac{1}{- 4 ( x + \frac{5}{2} ) ^{2} + 16} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int 1 d v

= 4 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= 4 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{2} (x + \frac{5}{2}))

Simplify

4 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{2} (x + \frac{5}{2})) : 2 arcsin (\frac{1}{2} (x + \frac{5}{2}))

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

Add a constant to the solution

= 2 arcsin (\frac{1}{2} (x + \frac{5}{2})) + C

f (x) = \frac{1}{x}, (- 1, - 1)

y^{^{'}} - 2 y = x^{2} + 5

\int \frac{1}{x ( x ^{7} + 1 ) ^{2}} d x

y = 2 \cdot x^{5} - 3 \cdot x \cdot e^{- x} - 10

x \to 2 lim (\frac{x - 4}{x ^{2} - 6})

What is the integral of 4/(sqrt(-4x^2-20x-9)) ?
The integral of 4/(sqrt(-4x^2-20x-9)) is 2arcsin(1/2 (x+5/2))+C