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

The integral of (2x+1)/(sqrt(x-2)) is 4/3 (x-2)^{3/2}+10sqrt(x-2)+C

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

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

\frac{4}{3} (x - 2)^{\frac{3}{2}} + 10 x - 2 + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = y (y^{3} - 1)

\int (e^{- x} - e^{5}) \cdot (e^{x} + e^{- 2}) d x

y^{^{'}} - y = 10 t e^{2 t}, y (0) = 1

f (x) = (- 25 x^{2} + 36)^{2}

\int_{0}^{2 π} sin^{2} (13 x) d x

What is the integral of (2x+1)/(sqrt(x-2)) ?
The integral of (2x+1)/(sqrt(x-2)) is 4/3 (x-2)^{3/2}+10sqrt(x-2)+C