What is the integral of (x+1)/(sqrt(3-2x-x^2)) ?

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

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integral of (x+1)/(sqrt(3-2x-x^2))

\int \frac{x + 1}{3 - 2 x - x ^{2}} d x

- - x^{2} - 2 x + 3 + C

Solution steps

\int \frac{x + 1}{3 - 2 x - x ^{2}} d x

Expand

\frac{x + 1}{3 - 2 x - x ^{2}} : \frac{x}{3 - 2 x - x ^{2}} + \frac{1}{3 - 2 x - x ^{2}}

= \int \frac{x}{3 - 2 x - x ^{2}} + \frac{1}{3 - 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}{3 - 2 x - x ^{2}} d x + \int \frac{1}{3 - 2 x - x ^{2}} d x

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

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

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

Simplify

= - - x^{2} - 2 x + 3

Add a constant to the solution

= - - x^{2} - 2 x + 3 + C

\frac{\partial}{\partial y} (e^{x y} tan (y))

\frac{d}{d x} (1 + \frac{5 e ^{- 2 x^{2}}}{2 - ∣ x ∣})

y = cos (sin (tan (7 x)))

\int_{0}^{8} \frac{1}{( 8 - x )} d x

\int 2 e^{- 0.8 x} d x

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