What is the integral of (6-2x)/(sqrt(8-4x-4x^2)) ?

The integral of (6-2x)/(sqrt(8-4x-4x^2)) is 7/2 arcsin(2/3 x+1/3)+sqrt(-x^2-x+2)+C

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integral of (6-2x)/(sqrt(8-4x-4x^2))

\int \frac{6 - 2 x}{8 - 4 x - 4 x ^{2}} d x

\frac{7}{2} arcsin (\frac{2}{3} x + \frac{1}{3}) + - x^{2} - x + 2 + C

Solution steps

\int \frac{6 - 2 x}{8 - 4 x - 4 x ^{2}} d x

Expand

\frac{6 - 2 x}{8 - 4 x - 4 x ^{2}} : \frac{3}{2 - x - x ^{2}} - \frac{x}{2 - x - x ^{2}}

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

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

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

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

Simplify

3 arcsin (\frac{2}{3} x + \frac{1}{3}) - (- - x^{2} - x + 2 - \frac{1}{2} arcsin (\frac{2}{3} x + \frac{1}{3})) : \frac{7}{2} arcsin (\frac{2}{3} x + \frac{1}{3}) + - x^{2} - x + 2

= \frac{7}{2} arcsin (\frac{2}{3} x + \frac{1}{3}) + - x^{2} - x + 2

Add a constant to the solution

= \frac{7}{2} arcsin (\frac{2}{3} x + \frac{1}{3}) + - x^{2} - x + 2 + C

\frac{d}{d z} (sin (y z))

\int cot (\frac{x}{2}) d x

f (- 2) = 3 x - 7

f (x) = sin (π x)

\frac{d}{d y} (x sin (y) + y cos (x))

What is the integral of (6-2x)/(sqrt(8-4x-4x^2)) ?
The integral of (6-2x)/(sqrt(8-4x-4x^2)) is 7/2 arcsin(2/3 x+1/3)+sqrt(-x^2-x+2)+C