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

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

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

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

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

Solution steps

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

Complete the square

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

= \int \frac{x}{- ( x - \frac{1}{2} ) ^{2} + \frac{9}{4}} d x

Apply u-substitution

: \int \frac{2 u + 1}{- 4 u ^{2} + 9} d u

= \int \frac{2 u + 1}{- 4 u ^{2} + 9} d u

Expand

\frac{2 u + 1}{- 4 u ^{2} + 9} : \frac{2 u}{- 4 u ^{2} + 9} + \frac{1}{- 4 u ^{2} + 9}

= \int \frac{2 u}{- 4 u ^{2} + 9} + \frac{1}{- 4 u ^{2} + 9} 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

= \int \frac{2 u}{- 4 u ^{2} + 9} d u + \int \frac{1}{- 4 u ^{2} + 9} d u

\int \frac{2 u}{- 4 u ^{2} + 9} d u = - \frac{1}{2} - 4 u^{2} + 9

\int \frac{1}{- 4 u ^{2} + 9} d u = \frac{1}{2} arcsin (\frac{2}{3} u)

= - \frac{1}{2} - 4 u^{2} + 9 + \frac{1}{2} arcsin (\frac{2}{3} u)

Substitute back

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

= - \frac{1}{2} - 4 (x - \frac{1}{2})^{2} + 9 + \frac{1}{2} arcsin (\frac{2}{3} (x - \frac{1}{2}))

Simplify

- \frac{1}{2} - 4 (x - \frac{1}{2})^{2} + 9 + \frac{1}{2} arcsin (\frac{2}{3} (x - \frac{1}{2})) : - - x^{2} + x + 2 + \frac{1}{2} arcsin (\frac{2}{3} x - \frac{1}{3})

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

Add a constant to the solution

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

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