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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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