What is the integral of (12)/(sqrt(3-8x-x^2)) ?

The integral of (12)/(sqrt(3-8x-x^2)) is 12arcsin(1/(sqrt(19))(x+4))+C

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integral of (12)/(sqrt(3-8x-x^2))

\int \frac{12}{3 - 8 x - x ^{2}} d x

12 arcsin (\frac{1}{19} (x + 4)) + C

Solution steps

\int \frac{12}{3 - 8 x - x ^{2}} d x

Take the constant out:

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

= 12 \cdot \int \frac{1}{3 - 8 x - x ^{2}} d x

Complete the square

3 - 8 x - x^{2} : - (x + 4)^{2} + 19

= 12 \cdot \int \frac{1}{- ( x + 4 ) ^{2} + 19} d x

Apply u-substitution

= 12 \cdot \int \frac{1}{- u ^{2} + 19} d u

Apply Trigonometric Substitution

= 12 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 12 \cdot 1 \cdot v

Substitute back

= 12 \cdot 1 \cdot arcsin (\frac{1}{19} (x + 4))

Simplify

= 12 arcsin (\frac{1}{19} (x + 4))

Add a constant to the solution

= 12 arcsin (\frac{1}{19} (x + 4)) + C

What is the integral of (12)/(sqrt(3-8x-x^2)) ?
The integral of (12)/(sqrt(3-8x-x^2)) is 12arcsin(1/(sqrt(19))(x+4))+C