What is the integral from 0 to 1 of (37)/(4y-1) ?

The integral from 0 to 1 of (37)/(4y-1) is diverges

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integral from 0 to 1 of (37)/(4y-1)

\int_{0}^{1} \frac{37}{4 y - 1} d y

diverges

Solution steps

\int_{0}^{1} \frac{37}{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{37}{4 y - 1} d y + \int_{\frac{1}{4}}^{1} \frac{37}{4 y - 1} d y

\int_{0}^{\frac{1}{4}} \frac{37}{4 y - 1} d y =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{3} arctan (x) d x

\frac{\partial}{\partial x} (- sin (x) - sin (2 y))

d er i v a t i v e f (x) = \frac{x ^{2}}{7 + x}

x y^{^{'}} - 2 y = 2 y x^{2}

x \to 0 lim (x \cdot cot (2 x))

What is the integral from 0 to 1 of (37)/(4y-1) ?
The integral from 0 to 1 of (37)/(4y-1) is diverges