What is the integral of (6x-5)/(sqrt(2-3x)) ?

The integral of (6x-5)/(sqrt(2-3x)) is 4/9 (2-3x)^{3/2}+2/3 sqrt(2-3x)+C

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integral of (6x-5)/(sqrt(2-3x))

\int \frac{6 x - 5}{2 - 3 x} d x

\frac{4}{9} (2 - 3 x)^{\frac{3}{2}} + \frac{2}{3} 2 - 3 x + C

Solution steps

\int \frac{6 x - 5}{2 - 3 x} d x

Expand

\frac{6 x - 5}{2 - 3 x} : \frac{6 x}{2 - 3 x} - \frac{5}{2 - 3 x}

= \int \frac{6 x}{2 - 3 x} - \frac{5}{2 - 3 x} 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{6 x}{2 - 3 x} d x - \int \frac{5}{2 - 3 x} d x

\int \frac{6 x}{2 - 3 x} d x = \frac{4}{3} (\frac{1}{3} (2 - 3 x)^{\frac{3}{2}} - 2 2 - 3 x)

\int \frac{5}{2 - 3 x} d x = - \frac{10}{3} 2 - 3 x

= \frac{4}{3} (\frac{1}{3} (2 - 3 x)^{\frac{3}{2}} - 2 2 - 3 x) - (- \frac{10}{3} 2 - 3 x)

Simplify

\frac{4}{3} (\frac{1}{3} (2 - 3 x)^{\frac{3}{2}} - 2 2 - 3 x) - (- \frac{10}{3} 2 - 3 x) : \frac{4}{9} (2 - 3 x)^{\frac{3}{2}} + \frac{2}{3} 2 - 3 x

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

Add a constant to the solution

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

What is the integral of (6x-5)/(sqrt(2-3x)) ?
The integral of (6x-5)/(sqrt(2-3x)) is 4/9 (2-3x)^{3/2}+2/3 sqrt(2-3x)+C