What is the integral of x/(sqrt(-9x^2+24x+20)) ?

The integral of x/(sqrt(-9x^2+24x+20)) is 1/9 (-sqrt(20+24x-9x^2)+4arcsin(1/2 x-2/3))+C

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integral of x/(sqrt(-9x^2+24x+20))

\int \frac{x}{- 9 x ^{2} + 24 x + 20} d x

\frac{1}{9} (- 20 + 24 x - 9 x^{2} + 4 arcsin (\frac{1}{2} x - \frac{2}{3})) + C

Solution steps

\int \frac{x}{- 9 x ^{2} + 24 x + 20} d x

Complete the square

- 9 x^{2} + 24 x + 20 : - 9 (x - \frac{4}{3})^{2} + 36

= \int \frac{x}{- 9 ( x - \frac{4}{3} ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{3 u + 4}{9 - u ^{2} + 4} d u

= \int \frac{3 u + 4}{9 - u ^{2} + 4} 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 + 4}{- u ^{2} + 4} d u

Expand

\frac{3 u + 4}{- u ^{2} + 4} : \frac{3 u}{- u ^{2} + 4} + \frac{4}{- u ^{2} + 4}

= \frac{1}{9} \cdot \int \frac{3 u}{- u ^{2} + 4} + \frac{4}{- u ^{2} + 4} 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} + 4} d u + \int \frac{4}{- u ^{2} + 4} d u)

\int \frac{3 u}{- u ^{2} + 4} d u = - 3 - u^{2} + 4

\int \frac{4}{- u ^{2} + 4} d u = 4 arcsin (\frac{1}{2} u)

= \frac{1}{9} (- 3 - u^{2} + 4 + 4 arcsin (\frac{1}{2} u))

Substitute back

u = x - \frac{4}{3}

= \frac{1}{9} - 3 - (x - \frac{4}{3})^{2} + 4 + 4 arcsin (\frac{1}{2} (x - \frac{4}{3}))

Simplify

\frac{1}{9} - 3 - (x - \frac{4}{3})^{2} + 4 + 4 arcsin (\frac{1}{2} (x - \frac{4}{3})) : \frac{1}{9} (- 20 + 24 x - 9 x^{2} + 4 arcsin (\frac{1}{2} x - \frac{2}{3}))

= \frac{1}{9} (- 20 + 24 x - 9 x^{2} + 4 arcsin (\frac{1}{2} x - \frac{2}{3}))

Add a constant to the solution

= \frac{1}{9} (- 20 + 24 x - 9 x^{2} + 4 arcsin (\frac{1}{2} x - \frac{2}{3})) + C

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What is the integral of x/(sqrt(-9x^2+24x+20)) ?
The integral of x/(sqrt(-9x^2+24x+20)) is 1/9 (-sqrt(20+24x-9x^2)+4arcsin(1/2 x-2/3))+C