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

The integral of x/(sqrt(5+x-x^2)) is -sqrt(-x^2+x+5)+1/2 arcsin(1/(sqrt(21))(2x-1))+C

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integral of x/(sqrt(5+x-x^2))

\int \frac{x}{5 + x - x ^{2}} d x

- - x^{2} + x + 5 + \frac{1}{2} arcsin (\frac{1}{21} (2 x - 1)) + C

Solution steps

\int \frac{x}{5 + x - x ^{2}} d x

Complete the square

5 + x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{21}{4}

= \int \frac{x}{- ( x - \frac{1}{2} ) ^{2} + \frac{21}{4}} d x

Apply u-substitution

: \int \frac{2 u + 1}{- 4 u ^{2} + 21} d u

= \int \frac{2 u + 1}{- 4 u ^{2} + 21} d u

Expand

\frac{2 u + 1}{- 4 u ^{2} + 21} : \frac{2 u}{- 4 u ^{2} + 21} + \frac{1}{- 4 u ^{2} + 21}

= \int \frac{2 u}{- 4 u ^{2} + 21} + \frac{1}{- 4 u ^{2} + 21} 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} + 21} d u + \int \frac{1}{- 4 u ^{2} + 21} d u

\int \frac{2 u}{- 4 u ^{2} + 21} d u = - \frac{1}{2} - 4 u^{2} + 21

\int \frac{1}{- 4 u ^{2} + 21} d u = \frac{1}{2} arcsin (\frac{2}{21} u)

= - \frac{1}{2} - 4 u^{2} + 21 + \frac{1}{2} arcsin (\frac{2}{21} u)

Substitute back

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

= - \frac{1}{2} - 4 (x - \frac{1}{2})^{2} + 21 + \frac{1}{2} arcsin (\frac{2}{21} (x - \frac{1}{2}))

Simplify

- \frac{1}{2} - 4 (x - \frac{1}{2})^{2} + 21 + \frac{1}{2} arcsin (\frac{2}{21} (x - \frac{1}{2})) : - - x^{2} + x + 5 + \frac{1}{2} arcsin (\frac{1}{21} (2 x - 1))

= - - x^{2} + x + 5 + \frac{1}{2} arcsin (\frac{1}{21} (2 x - 1))

Add a constant to the solution

= - - x^{2} + x + 5 + \frac{1}{2} arcsin (\frac{1}{21} (2 x - 1)) + C

\int x 2 x + 3 d x

\frac{\partial}{\partial x} (cos^{6} (x^{9} y^{5}))

\int \frac{( 2 x + 3 )}{x} d x

y^{^{''}} + 2 y^{^{'}} + 2 y = 8 e^{2 t}, y (0) = 0, y^{^{'}} (0) = 1

\frac{d y}{d t} = 0.5 y, y (2) = 100

What is the integral of x/(sqrt(5+x-x^2)) ?
The integral of x/(sqrt(5+x-x^2)) is -sqrt(-x^2+x+5)+1/2 arcsin(1/(sqrt(21))(2x-1))+C