What is the integral from 0 to 1 of 1/(4x-1) ?

The integral from 0 to 1 of 1/(4x-1) is diverges

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integral from 0 to 1 of 1/(4x-1)

\int_{0}^{1} \frac{1}{4 x - 1} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{4 x - 1} d x

Find undefined (singularity) points:

x = \frac{1}{4}

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}{4}} \frac{1}{4 x - 1} d x + \int_{\frac{1}{4}}^{1} \frac{1}{4 x - 1} d x

\int_{0}^{\frac{1}{4}} \frac{1}{4 x - 1} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int \frac{cos ( x )}{cos ( x ) + sin ( x )} d x

\int_{1}^{9} f (x) d x

x \to - 1 lim (arctan (\frac{x - 1}{x + 1}))

6 x^{3}

(5 - 7 x)^{3}

What is the integral from 0 to 1 of 1/(4x-1) ?
The integral from 0 to 1 of 1/(4x-1) is diverges