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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u}{- u ^{2} + 2 u + 35} d u

= \int \frac{u}{- u ^{2} + 2 u + 35} d u

Complete the square

- u^{2} + 2 u + 35 : - (u - 1)^{2} + 36

= \int \frac{u}{- ( u - 1 ) ^{2} + 36} d u

Apply u-substitution

: \int \frac{v + 1}{- v ^{2} + 36} d v

= \int \frac{v + 1}{- v ^{2} + 36} d v

Expand

\frac{v + 1}{- v ^{2} + 36} : \frac{v}{- v ^{2} + 36} + \frac{1}{- v ^{2} + 36}

= \int \frac{v}{- v ^{2} + 36} + \frac{1}{- v ^{2} + 36} d v

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{v}{- v ^{2} + 36} d v + \int \frac{1}{- v ^{2} + 36} d v

\int \frac{v}{- v ^{2} + 36} d v = - - v^{2} + 36

\int \frac{1}{- v ^{2} + 36} d v = arcsin (\frac{1}{6} v)

= - - v^{2} + 36 + arcsin (\frac{1}{6} v)

Substitute back

= - - (3 x - 2 - 1)^{2} + 36 + arcsin (\frac{1}{6} (3 x - 2 - 1))

Simplify

- - (3 x - 2 - 1)^{2} + 36 + arcsin (\frac{1}{6} (3 x - 2 - 1)) : - 3 - x^{2} + 2 x + 3 + arcsin (\frac{1}{2} (x - 1))

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

Add a constant to the solution

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

\int 4 x^{2} cos (6 x) d x

\frac{d}{d x} (\frac{2 x ^{3}}{2 x ^{4} - 5 x ^{3} + x ^{2}})

f (x) = 4 e^{x}

x (x - 10)

\frac{d}{d x} (\frac{2 x ^{2}}{3 x - 1})

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