What is the integral of 6/(sqrt(3-2x-x^2)) ?

The integral of 6/(sqrt(3-2x-x^2)) is 6arcsin(1/2 (x+1))+C

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integral of 6/(sqrt(3-2x-x^2))

\int \frac{6}{3 - 2 x - x ^{2}} d x

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

Solution steps

\int \frac{6}{3 - 2 x - x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{3 - 2 x - x ^{2}} d x

Complete the square

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

= 6 \cdot \int \frac{1}{- ( x + 1 ) ^{2} + 4} d x

Apply u-substitution

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

= 6 \cdot \int \frac{1}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int 1 d v

= 6 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 6 \cdot 1 \cdot v

Substitute back

= 6 \cdot 1 \cdot arcsin (\frac{1}{2} (x + 1))

Simplify

= 6 arcsin (\frac{1}{2} (x + 1))

Add a constant to the solution

= 6 arcsin (\frac{1}{2} (x + 1)) + C

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What is the integral of 6/(sqrt(3-2x-x^2)) ?
The integral of 6/(sqrt(3-2x-x^2)) is 6arcsin(1/2 (x+1))+C