What is the integral of (3x-2)/(sqrt(1-9x^2+6x)) ?

The integral of (3x-2)/(sqrt(1-9x^2+6x)) is 1/3 (-sqrt(-9x^2+6x+1)-arcsin(1/(sqrt(2))(3x-1)))+C

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integral of (3x-2)/(sqrt(1-9x^2+6x))

\int \frac{3 x - 2}{1 - 9 x ^{2} + 6 x} d x

\frac{1}{3} (- - 9 x^{2} + 6 x + 1 - arcsin (\frac{1}{2} (3 x - 1))) + C

Solution steps

\int \frac{3 x - 2}{1 - 9 x ^{2} + 6 x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Complete the square

- u^{2} - 2 u + 1 : - (u + 1)^{2} + 2

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

Apply u-substitution

: \int \frac{v - 1}{- v ^{2} + 2} d v

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

Expand

\frac{v - 1}{- v ^{2} + 2} : \frac{v}{- v ^{2} + 2} - \frac{1}{- v ^{2} + 2}

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

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

\int \frac{v}{- v ^{2} + 2} d v = - - v^{2} + 2

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

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

Substitute back

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

Simplify

\frac{1}{3} (- - (3 x - 2 + 1)^{2} + 2 - arcsin (\frac{1}{2} (3 x - 2 + 1))) : \frac{1}{3} (- - 9 x^{2} + 6 x + 1 - arcsin (\frac{1}{2} (3 x - 1)))

= \frac{1}{3} (- - 9 x^{2} + 6 x + 1 - arcsin (\frac{1}{2} (3 x - 1)))

Add a constant to the solution

= \frac{1}{3} (- - 9 x^{2} + 6 x + 1 - arcsin (\frac{1}{2} (3 x - 1))) + C

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What is the integral of (3x-2)/(sqrt(1-9x^2+6x)) ?
The integral of (3x-2)/(sqrt(1-9x^2+6x)) is 1/3 (-sqrt(-9x^2+6x+1)-arcsin(1/(sqrt(2))(3x-1)))+C