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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

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