What is the integral of (1-x)/(sqrt(5-2x-x^2)) ?

The integral of (1-x)/(sqrt(5-2x-x^2)) is 2arcsin(1/(sqrt(6))(x+1))+sqrt(-x^2-2x+5)+C

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integral of (1-x)/(sqrt(5-2x-x^2))

\int \frac{1 - x}{5 - 2 x - x ^{2}} d x

2 arcsin (\frac{1}{6} (x + 1)) + - x^{2} - 2 x + 5 + C

Solution steps

\int \frac{1 - x}{5 - 2 x - x ^{2}} d x

Expand

\frac{1 - x}{5 - 2 x - x ^{2}} : \frac{1}{5 - 2 x - x ^{2}} - \frac{x}{5 - 2 x - x ^{2}}

= \int \frac{1}{5 - 2 x - x ^{2}} - \frac{x}{5 - 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{1}{5 - 2 x - x ^{2}} d x - \int \frac{x}{5 - 2 x - x ^{2}} d x

\int \frac{1}{5 - 2 x - x ^{2}} d x = arcsin (\frac{1}{6} (x + 1))

\int \frac{x}{5 - 2 x - x ^{2}} d x = - - x^{2} - 2 x + 5 - arcsin (\frac{1}{6} (x + 1))

= arcsin (\frac{1}{6} (x + 1)) - (- - x^{2} - 2 x + 5 - arcsin (\frac{1}{6} (x + 1)))

Simplify

= 2 arcsin (\frac{1}{6} (x + 1)) + - x^{2} - 2 x + 5

Add a constant to the solution

= 2 arcsin (\frac{1}{6} (x + 1)) + - x^{2} - 2 x + 5 + C

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What is the integral of (1-x)/(sqrt(5-2x-x^2)) ?
The integral of (1-x)/(sqrt(5-2x-x^2)) is 2arcsin(1/(sqrt(6))(x+1))+sqrt(-x^2-2x+5)+C