What is the integral of x/(sqrt(4-2x-x^2)) ?

The integral of x/(sqrt(4-2x-x^2)) is -sqrt(-x^2-2x+4)-arcsin(1/(sqrt(5))(x+1))+C

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integral of x/(sqrt(4-2x-x^2))

\int \frac{x}{4 - 2 x - x ^{2}} d x

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

Solution steps

\int \frac{x}{4 - 2 x - x ^{2}} d x

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x + 1

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

Expand

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

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

Add a constant to the solution

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

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What is the integral of x/(sqrt(4-2x-x^2)) ?
The integral of x/(sqrt(4-2x-x^2)) is -sqrt(-x^2-2x+4)-arcsin(1/(sqrt(5))(x+1))+C