What is the integral of 8/(sqrt(12-x^2-4x)) ?

The integral of 8/(sqrt(12-x^2-4x)) is 8arcsin(1/4 (x+2))+C

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integral of 8/(sqrt(12-x^2-4x))

\int \frac{8}{12 - x ^{2} - 4 x} d x

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

Solution steps

\int \frac{8}{12 - x ^{2} - 4 x} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{12 - x ^{2} - 4 x} d x

Complete the square

12 - x^{2} - 4 x : - (x + 2)^{2} + 16

= 8 \cdot \int \frac{1}{- ( x + 2 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 16} d u

= 8 \cdot \int \frac{1}{- u ^{2} + 16} d u

Apply Trigonometric Substitution

: \int 1 d v

= 8 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 8 \cdot 1 \cdot v

Substitute back

= 8 \cdot 1 \cdot arcsin (\frac{1}{4} (x + 2))

Simplify

= 8 arcsin (\frac{1}{4} (x + 2))

Add a constant to the solution

= 8 arcsin (\frac{1}{4} (x + 2)) + C

3 x^{2}

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

\frac{d}{d x} (ln^{2} (sin (2 x)))

10 - x

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) (y + 7 y^{3}))

What is the integral of 8/(sqrt(12-x^2-4x)) ?
The integral of 8/(sqrt(12-x^2-4x)) is 8arcsin(1/4 (x+2))+C