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

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

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

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

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

Solution steps

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

Complete the square

4 + 4 x - x^{2} : - (x - 2)^{2} + 8

= \int \frac{x}{- ( x - 2 ) ^{2} + 8} d x

Apply u-substitution

: \int \frac{u + 2}{- u ^{2} + 8} d u

= \int \frac{u + 2}{- u ^{2} + 8} d u

Apply Trigonometric Substitution

: \int 22 sin (v) + 2 d v

= \int 22 sin (v) + 2 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 22 sin (v) d v + \int 2 d v

\int 22 sin (v) d v = - 22 cos (v)

\int 2 d v = 2 v

= - 22 cos (v) + 2 v

Substitute back

= - 22 cos (arcsin (\frac{1}{2 2} (x - 2))) + 2 arcsin (\frac{1}{2 2} (x - 2))

Simplify

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

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

Add a constant to the solution

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

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