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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

Expand

\frac{2 u + 1}{- 4 u ^{2} + 5} : \frac{2 u}{- 4 u ^{2} + 5} + \frac{1}{- 4 u ^{2} + 5}

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

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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