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

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

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

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

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

Solution steps

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

Expand

\frac{2 + x}{4 - 2 x - x ^{2}} : \frac{2}{4 - 2 x - x ^{2}} + \frac{x}{4 - 2 x - x ^{2}}

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int (sin (y) + y sin (x) + \frac{1}{x}) d x

\frac{d}{d x} (x^{2} \cdot x)

\frac{s}{( s - 3 ) ( s - 6 ) ( s - 2 )}

\int_{0}^{0.4} \frac{( x ^{2} )}{4 - 25 x ^{2}} d x

\int (sin (x)) (cos (x)) d x

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