What is the integral of x/(sqrt(8x-x^2)) ?

The integral of x/(sqrt(8x-x^2)) is -sqrt(-x^2+8x)+4arcsin(1/4 x-1)+C

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integral of x/(sqrt(8x-x^2))

\int \frac{x}{8 x - x ^{2}} d x

- - x^{2} + 8 x + 4 arcsin (\frac{1}{4} x - 1) + C

Solution steps

\int \frac{x}{8 x - x ^{2}} d x

Complete the square

8 x - x^{2} : - (x - 4)^{2} + 16

= \int \frac{x}{- ( x - 4 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{u + 4}{- u ^{2} + 16} d u

= \int \frac{u + 4}{- u ^{2} + 16} d u

Apply Trigonometric Substitution

: \int 4 (sin (v) + 1) d v

= \int 4 (sin (v) + 1) d v

Take the constant out:

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

= 4 \cdot \int sin (v) + 1 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

= 4 (\int sin (v) d v + \int 1 d v)

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

\int 1 d v = v

= 4 (- cos (v) + v)

Substitute back

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

Simplify

4 (- cos (arcsin (\frac{1}{4} (x - 4))) + arcsin (\frac{1}{4} (x - 4))) : - - x^{2} + 8 x + 4 arcsin (\frac{1}{4} x - 1)

= - - x^{2} + 8 x + 4 arcsin (\frac{1}{4} x - 1)

Add a constant to the solution

= - - x^{2} + 8 x + 4 arcsin (\frac{1}{4} x - 1) + C

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What is the integral of x/(sqrt(8x-x^2)) ?
The integral of x/(sqrt(8x-x^2)) is -sqrt(-x^2+8x)+4arcsin(1/4 x-1)+C