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

The integral of (2x+1)/(sqrt(3x+2)) is 4/27 (3x+2)^{3/2}-2/9 sqrt(3x+2)+C

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

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

\frac{4}{27} (3 x + 2)^{\frac{3}{2}} - \frac{2}{9} 3 x + 2 + C

Solution steps

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

Expand

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

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

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

Simplify

\frac{4}{9} (\frac{1}{3} (3 x + 2)^{\frac{3}{2}} - 2 3 x + 2) + \frac{2}{3} 3 x + 2 : \frac{4}{27} (3 x + 2)^{\frac{3}{2}} - \frac{2}{9} 3 x + 2

= \frac{4}{27} (3 x + 2)^{\frac{3}{2}} - \frac{2}{9} 3 x + 2

Add a constant to the solution

= \frac{4}{27} (3 x + 2)^{\frac{3}{2}} - \frac{2}{9} 3 x + 2 + C

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What is the integral of (2x+1)/(sqrt(3x+2)) ?
The integral of (2x+1)/(sqrt(3x+2)) is 4/27 (3x+2)^{3/2}-2/9 sqrt(3x+2)+C