What is the integral from 0 to 1 of (62)/(4y-1) ?

The integral from 0 to 1 of (62)/(4y-1) is diverges

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integral from 0 to 1 of (62)/(4y-1)

\int_{0}^{1} \frac{62}{4 y - 1} d y

diverges

Solution steps

\int_{0}^{1} \frac{62}{4 y - 1} d y

Find undefined (singularity) points:

y = \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{62}{4 y - 1} d y + \int_{\frac{1}{4}}^{1} \frac{62}{4 y - 1} d y

\int_{0}^{\frac{1}{4}} \frac{62}{4 y - 1} d y =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (\frac{x}{1 + ln ( x )})

y = \frac{3 x}{sin ( x )}

x \to - 1 lim (6 - 3 x)

\frac{\partial}{\partial x} (y)

\int_{0}^{2} [(- x^{2} + 4 x) - x^{2}] d x

What is the integral from 0 to 1 of (62)/(4y-1) ?
The integral from 0 to 1 of (62)/(4y-1) is diverges