What is the integral of (2x-3)/(sqrt(-3x^2+4x-1)) ?

The integral of (2x-3)/(sqrt(-3x^2+4x-1)) is sqrt(3)(-2/9 sqrt(-9x^2+12x-3)-5/9 arcsin(3x-2))+C

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integral of (2x-3)/(sqrt(-3x^2+4x-1))

\int \frac{2 x - 3}{- 3 x ^{2} + 4 x - 1} d x

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

Solution steps

\int \frac{2 x - 3}{- 3 x ^{2} + 4 x - 1} d x

Complete the square

- 3 x^{2} + 4 x - 1 : - 3 (x - \frac{2}{3})^{2} + \frac{1}{3}

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{2 ( u + \frac{2}{3} ) - 3}{- 9 u ^{2} + 1} : \frac{2 u}{- 9 u ^{2} + 1} - \frac{5}{3 - 9 u ^{2} + 1}

= 3 \cdot \int \frac{2 u}{- 9 u ^{2} + 1} - \frac{5}{3 - 9 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

= 3 (\int \frac{2 u}{- 9 u ^{2} + 1} d u - \int \frac{5}{3 - 9 u ^{2} + 1} d u)

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

\int \frac{5}{3 - 9 u ^{2} + 1} d u = \frac{5}{9} arcsin (3 u)

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (2x-3)/(sqrt(-3x^2+4x-1)) ?
The integral of (2x-3)/(sqrt(-3x^2+4x-1)) is sqrt(3)(-2/9 sqrt(-9x^2+12x-3)-5/9 arcsin(3x-2))+C