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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x - 2

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

Expand

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

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

Add a constant to the solution

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

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