What is the integral of (2x-3)/(sqrt(3+4x-4x^2)) ?

The integral of (2x-3)/(sqrt(3+4x-4x^2)) is 1/2 (-sqrt(-4x^2+4x+3)-2arcsin(x-1/2))+C

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integral of (2x-3)/(sqrt(3+4x-4x^2))

\int \frac{2 x - 3}{3 + 4 x - 4 x ^{2}} d x

\frac{1}{2} (- - 4 x^{2} + 4 x + 3 - 2 arcsin (x - \frac{1}{2})) + C

Solution steps

\int \frac{2 x - 3}{3 + 4 x - 4 x ^{2}} d x

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Complete the square

- u^{2} - 4 u : - (u + 2)^{2} + 4

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

Apply u-substitution

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

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

Expand

\frac{v - 2}{- v ^{2} + 4} : \frac{v}{- v ^{2} + 4} - \frac{2}{- v ^{2} + 4}

= \frac{1}{2} \cdot \int \frac{v}{- v ^{2} + 4} - \frac{2}{- v ^{2} + 4} 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

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

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

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

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

Substitute back

= \frac{1}{2} (- - (2 x - 3 + 2)^{2} + 4 - 2 arcsin (\frac{1}{2} (2 x - 3 + 2)))

Simplify

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

= \frac{1}{2} (- - 4 x^{2} + 4 x + 3 - 2 arcsin (x - \frac{1}{2}))

Add a constant to the solution

= \frac{1}{2} (- - 4 x^{2} + 4 x + 3 - 2 arcsin (x - \frac{1}{2})) + C

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What is the integral of (2x-3)/(sqrt(3+4x-4x^2)) ?
The integral of (2x-3)/(sqrt(3+4x-4x^2)) is 1/2 (-sqrt(-4x^2+4x+3)-2arcsin(x-1/2))+C