What is the integral of (x+4)/(sqrt(6x-x^2)) ?

The integral of (x+4)/(sqrt(6x-x^2)) is -sqrt(-x^2+6x)+7arcsin(1/3 x-1)+C

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integral of (x+4)/(sqrt(6x-x^2))

\int \frac{x + 4}{6 x - x ^{2}} d x

- - x^{2} + 6 x + 7 arcsin (\frac{1}{3} x - 1) + C

Solution steps

\int \frac{x + 4}{6 x - x ^{2}} d x

Complete the square

6 x - x^{2} : - (x - 3)^{2} + 9

= \int \frac{x + 4}{- ( x - 3 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{u + 7}{- u ^{2} + 9} d u

= \int \frac{u + 7}{- u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int 3 sin (v) + 7 d v

= \int 3 sin (v) + 7 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 3 sin (v) d v + \int 7 d v

\int 3 sin (v) d v = - 3 cos (v)

\int 7 d v = 7 v

= - 3 cos (v) + 7 v

Substitute back

= - 3 cos (arcsin (\frac{1}{3} (x - 3))) + 7 arcsin (\frac{1}{3} (x - 3))

Simplify

- 3 cos (arcsin (\frac{1}{3} (x - 3))) + 7 arcsin (\frac{1}{3} (x - 3)) : - - x^{2} + 6 x + 7 arcsin (\frac{1}{3} x - 1)

= - - x^{2} + 6 x + 7 arcsin (\frac{1}{3} x - 1)

Add a constant to the solution

= - - x^{2} + 6 x + 7 arcsin (\frac{1}{3} x - 1) + C

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What is the integral of (x+4)/(sqrt(6x-x^2)) ?
The integral of (x+4)/(sqrt(6x-x^2)) is -sqrt(-x^2+6x)+7arcsin(1/3 x-1)+C