What is the integral of x/(sqrt(5+12x-9x^2)) ?

The integral of x/(sqrt(5+12x-9x^2)) is 1/9 (-sqrt(5+12x-9x^2)+2arcsin(x-2/3))+C

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integral of x/(sqrt(5+12x-9x^2))

\int \frac{x}{5 + 12 x - 9 x ^{2}} d x

\frac{1}{9} (- 5 + 12 x - 9 x^{2} + 2 arcsin (x - \frac{2}{3})) + C

Solution steps

\int \frac{x}{5 + 12 x - 9 x ^{2}} d x

Complete the square

5 + 12 x - 9 x^{2} : - 9 (x - \frac{2}{3})^{2} + 9

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

Apply u-substitution

: \int \frac{3 u + 2}{9 - u ^{2} + 1} d u

= \int \frac{3 u + 2}{9 - u ^{2} + 1} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{3 u + 2}{- u ^{2} + 1} d u

Expand

\frac{3 u + 2}{- u ^{2} + 1} : \frac{3 u}{- u ^{2} + 1} + \frac{2}{- u ^{2} + 1}

= \frac{1}{9} \cdot \int \frac{3 u}{- u ^{2} + 1} + \frac{2}{- 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

= \frac{1}{9} (\int \frac{3 u}{- u ^{2} + 1} d u + \int \frac{2}{- u ^{2} + 1} d u)

\int \frac{3 u}{- u ^{2} + 1} d u = - 3 - u^{2} + 1

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

= \frac{1}{9} (- 3 - u^{2} + 1 + 2 arcsin (u))

Substitute back

u = x - \frac{2}{3}

= \frac{1}{9} - 3 - (x - \frac{2}{3})^{2} + 1 + 2 arcsin (x - \frac{2}{3})

Simplify

\frac{1}{9} - 3 - (x - \frac{2}{3})^{2} + 1 + 2 arcsin (x - \frac{2}{3}) : \frac{1}{9} (- 5 + 12 x - 9 x^{2} + 2 arcsin (x - \frac{2}{3}))

= \frac{1}{9} (- 5 + 12 x - 9 x^{2} + 2 arcsin (x - \frac{2}{3}))

Add a constant to the solution

= \frac{1}{9} (- 5 + 12 x - 9 x^{2} + 2 arcsin (x - \frac{2}{3})) + C

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What is the integral of x/(sqrt(5+12x-9x^2)) ?
The integral of x/(sqrt(5+12x-9x^2)) is 1/9 (-sqrt(5+12x-9x^2)+2arcsin(x-2/3))+C