What is the integral of x/(sqrt(27+6x-x^2)) ?

The integral of x/(sqrt(27+6x-x^2)) is -sqrt(-x^2+6x+27)+3arcsin(1/6 (x-3))+C

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integral of x/(sqrt(27+6x-x^2))

\int \frac{x}{27 + 6 x - x ^{2}} d x

- - x^{2} + 6 x + 27 + 3 arcsin (\frac{1}{6} (x - 3)) + C

Solution steps

\int \frac{x}{27 + 6 x - x ^{2}} d x

Complete the square

27 + 6 x - x^{2} : - (x - 3)^{2} + 36

= \int \frac{x}{- ( x - 3 ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{u + 3}{- u ^{2} + 36} d u

= \int \frac{u + 3}{- u ^{2} + 36} d u

Apply Trigonometric Substitution

: \int 3 (2 sin (v) + 1) d v

= \int 3 (2 sin (v) + 1) d v

Take the constant out:

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

= 3 \cdot \int 2 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

= 3 (\int 2 sin (v) d v + \int 1 d v)

\int 2 sin (v) d v = - 2 cos (v)

\int 1 d v = v

= 3 (- 2 cos (v) + v)

Substitute back

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

Simplify

3 (- 2 cos (arcsin (\frac{1}{6} (x - 3))) + arcsin (\frac{1}{6} (x - 3))) : - - x^{2} + 6 x + 27 + 3 arcsin (\frac{1}{6} (x - 3))

= - - x^{2} + 6 x + 27 + 3 arcsin (\frac{1}{6} (x - 3))

Add a constant to the solution

= - - x^{2} + 6 x + 27 + 3 arcsin (\frac{1}{6} (x - 3)) + C

tan^{40} (sec (cos (x)))

\frac{4}{s ^{2} ( s + 1 )}

n \to \infty lim (\frac{3 ^{n + 1}}{( n ) !})

y^{^{'}} = \frac{x ^{3} - 2 y}{x}

\int t^{2} + e^{t} d t

What is the integral of x/(sqrt(27+6x-x^2)) ?
The integral of x/(sqrt(27+6x-x^2)) is -sqrt(-x^2+6x+27)+3arcsin(1/6 (x-3))+C