What is the integral of ((x+6))/(sqrt(5-4x-x^2)) ?

The integral of ((x+6))/(sqrt(5-4x-x^2)) is -sqrt(-x^2-4x+5)+4arcsin(1/3 (x+2))+C

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integral of ((x+6))/(sqrt(5-4x-x^2))

\int \frac{( x + 6 )}{5 - 4 x - x ^{2}} d x

- - x^{2} - 4 x + 5 + 4 arcsin (\frac{1}{3} (x + 2)) + C

Solution steps

\int \frac{x + 6}{5 - 4 x - x ^{2}} d x

Complete the square

5 - 4 x - x^{2} : - (x + 2)^{2} + 9

= \int \frac{x + 6}{- ( x + 2 ) ^{2} + 9} d x

Apply u-substitution

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

Apply Trigonometric Substitution

= \int 3 sin (v) + 4 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 sin (v) d v + \int 4 d v

\int 3 sin (v) d v = - 3 cos (v)

\int 4 d v = 4 v

= - 3 cos (v) + 4 v

Substitute back

= - 3 cos (arcsin (\frac{1}{3} (x + 2))) + 4 arcsin (\frac{1}{3} (x + 2))

Simplify

- 3 cos (arcsin (\frac{1}{3} (x + 2))) + 4 arcsin (\frac{1}{3} (x + 2)) : - - x^{2} - 4 x + 5 + 4 arcsin (\frac{1}{3} (x + 2))

= - - x^{2} - 4 x + 5 + 4 arcsin (\frac{1}{3} (x + 2))

Add a constant to the solution

= - - x^{2} - 4 x + 5 + 4 arcsin (\frac{1}{3} (x + 2)) + C

What is the integral of ((x+6))/(sqrt(5-4x-x^2)) ?
The integral of ((x+6))/(sqrt(5-4x-x^2)) is -sqrt(-x^2-4x+5)+4arcsin(1/3 (x+2))+C