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

The integral of (x-1)/(sqrt(3x-2)) is 2/27 (3x-2)^{3/2}-2/9 sqrt(3x-2)+C

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integral of (x-1)/(sqrt(3x-2))

\int \frac{x - 1}{3 x - 2} d x

\frac{2}{27} (3 x - 2)^{\frac{3}{2}} - \frac{2}{9} 3 x - 2 + C

Solution steps

\int \frac{x - 1}{3 x - 2} d x

Expand

\frac{x - 1}{3 x - 2} : \frac{x}{3 x - 2} - \frac{1}{3 x - 2}

= \int \frac{x}{3 x - 2} - \frac{1}{3 x - 2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{3 x - 2} d x - \int \frac{1}{3 x - 2} d x

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

\int \frac{1}{3 x - 2} d x = \frac{2}{3} 3 x - 2

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

Simplify

\frac{2}{9} (\frac{1}{3} (3 x - 2)^{\frac{3}{2}} + 2 3 x - 2) - \frac{2}{3} 3 x - 2 : \frac{2}{27} (3 x - 2)^{\frac{3}{2}} - \frac{2}{9} 3 x - 2

= \frac{2}{27} (3 x - 2)^{\frac{3}{2}} - \frac{2}{9} 3 x - 2

Add a constant to the solution

= \frac{2}{27} (3 x - 2)^{\frac{3}{2}} - \frac{2}{9} 3 x - 2 + C

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What is the integral of (x-1)/(sqrt(3x-2)) ?
The integral of (x-1)/(sqrt(3x-2)) is 2/27 (3x-2)^{3/2}-2/9 sqrt(3x-2)+C