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

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

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

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

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

Solution steps

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

Complete the square

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

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

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 = s + 1

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

Expand

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

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

Add a constant to the solution

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

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