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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int_{0}^{1} 13 \frac{1}{x ^{p}} d x

\frac{\partial}{\partial x} (5 x y e^{- y^{2}})

\int (\frac{15}{x ^{4}} + \frac{8}{x ^{5}}) d x

\frac{d}{d x} (\frac{sin ( 3 x )}{x})

6 y^{^{''}} + 8 y^{^{'}} + y = 0, y (- 2) = - 3, y^{^{'}} (- 2) = - 4

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