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

The integral of 4/(sqrt(-x^2-4x-1)) is 4arcsin(1/(sqrt(3))(x+2))+C

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

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

4 arcsin (\frac{1}{3} (x + 2)) + C

Solution steps

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

Complete the square

- x^{2} - 4 x - 1 : - (x + 2)^{2} + 3

= 4 \cdot \int \frac{1}{- ( x + 2 ) ^{2} + 3} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 3} d u

= 4 \cdot \int \frac{1}{- u ^{2} + 3} d u

Apply Trigonometric Substitution

: \int 1 d v

= 4 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 4 \cdot 1 \cdot v

Substitute back

= 4 \cdot 1 \cdot arcsin (\frac{1}{3} (x + 2))

Simplify

= 4 arcsin (\frac{1}{3} (x + 2))

Add a constant to the solution

= 4 arcsin (\frac{1}{3} (x + 2)) + C

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What is the integral of 4/(sqrt(-x^2-4x-1)) ?
The integral of 4/(sqrt(-x^2-4x-1)) is 4arcsin(1/(sqrt(3))(x+2))+C