What is the integral of x/(sqrt(7-x-x^2)) ?

The integral of x/(sqrt(7-x-x^2)) is -sqrt(-x^2-x+7)-1/2 arcsin(1/(sqrt(29))(2x+1))+C

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integral of x/(sqrt(7-x-x^2))

\int \frac{x}{7 - x - x ^{2}} d x

- - x^{2} - x + 7 - \frac{1}{2} arcsin (\frac{1}{29} (2 x + 1)) + C

Solution steps

\int \frac{x}{7 - x - x ^{2}} d x

Complete the square

7 - x - x^{2} : - (x + \frac{1}{2})^{2} + \frac{29}{4}

= \int \frac{x}{- ( x + \frac{1}{2} ) ^{2} + \frac{29}{4}} d x

Apply u-substitution

: \int \frac{2 u - 1}{- 4 u ^{2} + 29} d u

= \int \frac{2 u - 1}{- 4 u ^{2} + 29} d u

Expand

\frac{2 u - 1}{- 4 u ^{2} + 29} : \frac{2 u}{- 4 u ^{2} + 29} - \frac{1}{- 4 u ^{2} + 29}

= \int \frac{2 u}{- 4 u ^{2} + 29} - \frac{1}{- 4 u ^{2} + 29} 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} + 29} d u - \int \frac{1}{- 4 u ^{2} + 29} d u

\int \frac{2 u}{- 4 u ^{2} + 29} d u = - \frac{1}{2} - 4 u^{2} + 29

\int \frac{1}{- 4 u ^{2} + 29} d u = \frac{1}{2} arcsin (\frac{2}{29} u)

= - \frac{1}{2} - 4 u^{2} + 29 - \frac{1}{2} arcsin (\frac{2}{29} u)

Substitute back

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

= - \frac{1}{2} - 4 (x + \frac{1}{2})^{2} + 29 - \frac{1}{2} arcsin (\frac{2}{29} (x + \frac{1}{2}))

Simplify

- \frac{1}{2} - 4 (x + \frac{1}{2})^{2} + 29 - \frac{1}{2} arcsin (\frac{2}{29} (x + \frac{1}{2})) : - - x^{2} - x + 7 - \frac{1}{2} arcsin (\frac{1}{29} (2 x + 1))

= - - x^{2} - x + 7 - \frac{1}{2} arcsin (\frac{1}{29} (2 x + 1))

Add a constant to the solution

= - - x^{2} - x + 7 - \frac{1}{2} arcsin (\frac{1}{29} (2 x + 1)) + C

\int \frac{1}{169 - 16 x ^{2}} d x

\frac{d}{d x} (e^{x a})

t y^{^{''}} + 4 y^{^{'}} = t^{2}

f f^{^{'}} (xx) = x x^{3} - x x^{2} - 6 xx

\frac{d}{d x} (e^{c o s h (x)} cos^{2} (x))

What is the integral of x/(sqrt(7-x-x^2)) ?
The integral of x/(sqrt(7-x-x^2)) is -sqrt(-x^2-x+7)-1/2 arcsin(1/(sqrt(29))(2x+1))+C