What is the integral of 1/(sqrt(8-4x-4x^2)) ?

The integral of 1/(sqrt(8-4x-4x^2)) is 1/2 arcsin(2/3 x+1/3)+C

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integral of 1/(sqrt(8-4x-4x^2))

\int \frac{1}{8 - 4 x - 4 x ^{2}} d x

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

Solution steps

\int \frac{1}{8 - 4 x - 4 x ^{2}} d x

Complete the square

8 - 4 x - 4 x^{2} : - 4 (x + \frac{1}{2})^{2} + 9

= \int \frac{1}{- 4 ( x + \frac{1}{2} ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{- 4 u ^{2} + 9} d u

= \int \frac{1}{- 4 u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} + 2 y = 3 e^{t}

\int_{0}^{2} \frac{x}{x ^{2} + 4} d x

x \to - 3 - lim (\frac{1}{x ^{2} - 9})

\int \frac{1}{4 x ( 6 + x )} d x

\frac{2000 t}{4 t + 75}

What is the integral of 1/(sqrt(8-4x-4x^2)) ?
The integral of 1/(sqrt(8-4x-4x^2)) is 1/2 arcsin(2/3 x+1/3)+C