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

The integral of x/(sqrt(6x-x^2-4)) is -sqrt(5)sqrt(1/5 (-x^2+6x-4))+3arcsin(1/(sqrt(5))(x-3))+C

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

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

- 5 \frac{1}{5} (- x^{2} + 6 x - 4) + 3 arcsin (\frac{1}{5} (x - 3)) + C

Solution steps

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

Complete the square

6 x - x^{2} - 4 : - (x - 3)^{2} + 5

= \int \frac{x}{- ( x - 3 ) ^{2} + 5} d x

Apply u-substitution

: \int \frac{u + 3}{- u ^{2} + 5} d u

= \int \frac{u + 3}{- u ^{2} + 5} d u

Apply Trigonometric Substitution

: \int 5 sin (v) + 3 d v

= \int 5 sin (v) + 3 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 5 sin (v) d v + \int 3 d v

\int 5 sin (v) d v = - 5 cos (v)

\int 3 d v = 3 v

= - 5 cos (v) + 3 v

Substitute back

= - 5 cos (arcsin (\frac{1}{5} (x - 3))) + 3 arcsin (\frac{1}{5} (x - 3))

Simplify

- 5 cos (arcsin (\frac{1}{5} (x - 3))) + 3 arcsin (\frac{1}{5} (x - 3)) : - 5 \frac{1}{5} (- x^{2} + 6 x - 4) + 3 arcsin (\frac{1}{5} (x - 3))

= - 5 \frac{1}{5} (- x^{2} + 6 x - 4) + 3 arcsin (\frac{1}{5} (x - 3))

Add a constant to the solution

= - 5 \frac{1}{5} (- x^{2} + 6 x - 4) + 3 arcsin (\frac{1}{5} (x - 3)) + C

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What is the integral of x/(sqrt(6x-x^2-4)) ?
The integral of x/(sqrt(6x-x^2-4)) is -sqrt(5)sqrt(1/5 (-x^2+6x-4))+3arcsin(1/(sqrt(5))(x-3))+C