What is the integral of (2x-8)/(sqrt(1-x-x^2)) ?

The integral of (2x-8)/(sqrt(1-x-x^2)) is -2sqrt(-x^2-x+1)-9arcsin(1/(sqrt(5))(2x+1))+C

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integral of (2x-8)/(sqrt(1-x-x^2))

\int \frac{2 x - 8}{1 - x - x ^{2}} d x

- 2 - x^{2} - x + 1 - 9 arcsin (\frac{1}{5} (2 x + 1)) + C

Solution steps

\int \frac{2 x - 8}{1 - x - x ^{2}} d x

Complete the square

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

= \int \frac{2 x - 8}{- ( x + \frac{1}{2} ) ^{2} + \frac{5}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u - 9 )}{- 4 u ^{2} + 5} d u

= \int \frac{2 ( 2 u - 9 )}{- 4 u ^{2} + 5} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{2 u - 9}{- 4 u ^{2} + 5} d u

Expand

\frac{2 u - 9}{- 4 u ^{2} + 5} : \frac{2 u}{- 4 u ^{2} + 5} - \frac{9}{- 4 u ^{2} + 5}

= 2 \cdot \int \frac{2 u}{- 4 u ^{2} + 5} - \frac{9}{- 4 u ^{2} + 5} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{2 u}{- 4 u ^{2} + 5} d u - \int \frac{9}{- 4 u ^{2} + 5} d u)

\int \frac{2 u}{- 4 u ^{2} + 5} d u = - \frac{1}{2} - 4 u^{2} + 5

\int \frac{9}{- 4 u ^{2} + 5} d u = \frac{9}{2} arcsin (\frac{2}{5} u)

= 2 (- \frac{1}{2} - 4 u^{2} + 5 - \frac{9}{2} arcsin (\frac{2}{5} u))

Substitute back

u = x + \frac{1}{2}

= 2 - \frac{1}{2} - 4 (x + \frac{1}{2})^{2} + 5 - \frac{9}{2} arcsin (\frac{2}{5} (x + \frac{1}{2}))

Simplify

2 - \frac{1}{2} - 4 (x + \frac{1}{2})^{2} + 5 - \frac{9}{2} arcsin (\frac{2}{5} (x + \frac{1}{2})) : - 2 - x^{2} - x + 1 - 9 arcsin (\frac{1}{5} (2 x + 1))

= - 2 - x^{2} - x + 1 - 9 arcsin (\frac{1}{5} (2 x + 1))

Add a constant to the solution

= - 2 - x^{2} - x + 1 - 9 arcsin (\frac{1}{5} (2 x + 1)) + C

\frac{\partial}{\partial y} (x ln (x y - 5))

\int_{0}^{9} (\frac{1}{3} x - 2) d x

(5 + x) y^{^{'}} - y = 0

\int u - 1 d u

\int \frac{tan ( ln ( 4 x + 3 ) )}{4 x + 3} d x

What is the integral of (2x-8)/(sqrt(1-x-x^2)) ?
The integral of (2x-8)/(sqrt(1-x-x^2)) is -2sqrt(-x^2-x+1)-9arcsin(1/(sqrt(5))(2x+1))+C