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

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

Apply Trigonometric Substitution

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

= \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

\frac{d}{d t} (ln (cos (\frac{π}{2} - \frac{t}{2})))

\int (x + 2) e^{4 - 3 x} d x

n = 1 \sum \infty \frac{3 ^{n}}{n}

(t - 1) cos (4 (t - 1))

(x - 1) y^{^{'}} + x y = 6 e^{x}

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