What is the integral of x/(sqrt(3-2x-x^2)) ?

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

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integral of x/(sqrt(3-2x-x^2))

\int \frac{x}{3 - 2 x - x ^{2}} d x

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

Solution steps

\int \frac{x}{3 - 2 x - x ^{2}} d x

Complete the square

3 - 2 x - x^{2} : - (x + 1)^{2} + 4

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x + 1

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

Expand

- (x + 1)^{2} + 4 : - x^{2} - 2 x + 3

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

Add a constant to the solution

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

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

\int_{2}^{\frac{1}{x}} arctan (t) d t

(cos (π))^{^{'}}

\int_{- 5}^{0} (4 + 25 - x^{2}) d x

x \to 8 lim (\frac{x + 9}{x - 8})

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