What is the integral from 0 to 6 of 1/(sqrt(x-3)) ?

The integral from 0 to 6 of 1/(sqrt(x-3)) is -2sqrt(3)i+2sqrt(3)

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integral from 0 to 6 of 1/(sqrt(x-3))

\int_{0}^{6} \frac{1}{x - 3} d x

- 23 i + 23

Solution steps

\int_{0}^{6} \frac{1}{x - 3} d x

Find undefined (singularity) points:

x = 3

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{3} \frac{1}{x - 3} d x + \int_{3}^{6} \frac{1}{x - 3} d x

\int_{0}^{3} \frac{1}{x - 3} d x = - 23 i

\int_{3}^{6} \frac{1}{x - 3} d x = 23

= - 23 i + 23

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

\int cos (\frac{nπ x}{2 π}) d x

x \to 2 - lim (\frac{3}{2 - x})

f^{^{'}} (x) = 2 x^{2} + x - 1

\int 64 - x^{2} d x

What is the integral from 0 to 6 of 1/(sqrt(x-3)) ?
The integral from 0 to 6 of 1/(sqrt(x-3)) is -2sqrt(3)i+2sqrt(3)