What is the integral from 0 to 1 of 1/(2x-1) ?

The integral from 0 to 1 of 1/(2x-1) is diverges

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integral from 0 to 1 of 1/(2x-1)

\int_{0}^{1} \frac{1}{2 x - 1} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{2 x - 1} d x

Find undefined (singularity) points:

x = \frac{1}{2}

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}^{\frac{1}{2}} \frac{1}{2 x - 1} d x + \int_{\frac{1}{2}}^{1} \frac{1}{2 x - 1} d x

\int_{0}^{\frac{1}{2}} \frac{1}{2 x - 1} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{\partial}{\partial x} (sec (θ))

\frac{d}{d x} (\frac{( - 2 x ^{3} - 1 )}{( x ^{3} - 1 ) ^{2}})

\int_{0}^{x} 1 + \frac{9}{4} t d t

h \to 0 lim (\frac{2. 8 ^{h} - 1}{h})

\frac{d}{d x} (2^{- 3 x})

What is the integral from 0 to 1 of 1/(2x-1) ?
The integral from 0 to 1 of 1/(2x-1) is diverges