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

The integral from 2 to 3 of (2x-3)/(sqrt(4x-x^2)) is 4-2sqrt(3)+pi/6

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

\int_{2}^{3} \frac{2 x - 3}{4 x - x ^{2}} d x

4 - 23 + \frac{π}{6}

Decimal Notation

1.05949 \dots

Solution steps

\int_{2}^{3} \frac{2 x - 3}{4 x - x ^{2}} d x

Complete the square

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

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

Apply u-substitution

: \int_{0}^{1} \frac{2 u + 1}{- u ^{2} + 4} d u

= \int_{0}^{1} \frac{2 u + 1}{- u ^{2} + 4} d u

Expand

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

= \int_{0}^{1} \frac{2 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_{0}^{1} \frac{2 u}{- u ^{2} + 4} d u + \int_{0}^{1} \frac{1}{- u ^{2} + 4} d u

\int_{0}^{1} \frac{2 u}{- u ^{2} + 4} d u = 4 - 23

\int_{0}^{1} \frac{1}{- u ^{2} + 4} d u = \frac{π}{6}

= 4 - 23 + \frac{π}{6}

\int sin (3 x - 2) e^{2 x} d x

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

y^{^{''}} + 2 y = sin (2 x)

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

\frac{d}{d x} (\frac{2 x}{2})

What is the integral from 2 to 3 of (2x-3)/(sqrt(4x-x^2)) ?
The integral from 2 to 3 of (2x-3)/(sqrt(4x-x^2)) is 4-2sqrt(3)+pi/6