What is the integral of (x+3)/(sqrt(5-4x-x^2)) ?

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

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integral of (x+3)/(sqrt(5-4x-x^2))

\int \frac{x + 3}{5 - 4 x - x ^{2}} d x

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

Solution steps

\int \frac{x + 3}{5 - 4 x - x ^{2}} d x

Complete the square

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

= \int \frac{x + 3}{- ( x + 2 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{u + 1}{- u ^{2} + 9} d u

= \int \frac{u + 1}{- u ^{2} + 9} d u

Expand

\frac{u + 1}{- u ^{2} + 9} : \frac{u}{- u ^{2} + 9} + \frac{1}{- u ^{2} + 9}

= \int \frac{u}{- u ^{2} + 9} + \frac{1}{- u ^{2} + 9} d u

Apply the Sum Rule:

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

= \int \frac{u}{- u ^{2} + 9} d u + \int \frac{1}{- u ^{2} + 9} d u

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

\int \frac{1}{- u ^{2} + 9} d u = arcsin (\frac{1}{3} u)

= - - u^{2} + 9 + arcsin (\frac{1}{3} u)

Substitute back

u = x + 2

= - - (x + 2)^{2} + 9 + arcsin (\frac{1}{3} (x + 2))

Expand

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

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

Add a constant to the solution

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

\frac{d ^{2}}{d x ^{2}} (\frac{( x + b ) ( x - a )}{( x - b )})

(cos (x) sin (x) - x y^{2} 2) d x + y (1 - x^{2}) d y = 0

\frac{3 x}{2 - cot ( x )}

\frac{d}{d z} (e^{3 x + 4 y} cos (5 z))

\frac{( e ^{x} )}{3 x ^{4}}

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