What is the integral from 0 to 1 of (17)/(4x-1) ?

The integral from 0 to 1 of (17)/(4x-1) is diverges

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integral from 0 to 1 of (17)/(4x-1)

\int_{0}^{1} \frac{17}{4 x - 1} d x

diverges

Solution steps

\int_{0}^{1} \frac{17}{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{17}{4 x - 1} d x + \int_{\frac{1}{4}}^{1} \frac{17}{4 x - 1} d x

\int_{0}^{\frac{1}{4}} \frac{17}{4 x - 1} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

ln (1 - 4 y)

\frac{\partial}{\partial x} (cos (8 x) sin (4 y))

\frac{1}{( 2 x ^{3} + 3 x + 5 ) ^{\frac{3}{4}}}

\int_{1}^{e} (ln (x))^{2} d x

\frac{d}{d x} (- sec (x))

What is the integral from 0 to 1 of (17)/(4x-1) ?
The integral from 0 to 1 of (17)/(4x-1) is diverges