What is the integral of 6/(sqrt(6x-x^2-5)) ?

The integral of 6/(sqrt(6x-x^2-5)) is 6arcsin(1/2 (x-3))+C

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integral of 6/(sqrt(6x-x^2-5))

\int \frac{6}{6 x - x ^{2} - 5} d x

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

Solution steps

\int \frac{6}{6 x - x ^{2} - 5} 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}{6 x - x ^{2} - 5} d x

Complete the square

6 x - x^{2} - 5 : - (x - 3)^{2} + 4

= 6 \cdot \int \frac{1}{- ( x - 3 ) ^{2} + 4} d x

Apply u-substitution

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

Apply Trigonometric Substitution

= 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 - 3))

Simplify

= 6 arcsin (\frac{1}{2} (x - 3))

Add a constant to the solution

= 6 arcsin (\frac{1}{2} (x - 3)) + C

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What is the integral of 6/(sqrt(6x-x^2-5)) ?
The integral of 6/(sqrt(6x-x^2-5)) is 6arcsin(1/2 (x-3))+C