What is the integral from 0 to 1 of (33)/(4y-1) ?

The integral from 0 to 1 of (33)/(4y-1) is diverges

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integral from 0 to 1 of (33)/(4y-1)

\int_{0}^{1} \frac{33}{4 y - 1} d y

diverges

Solution steps

\int_{0}^{1} \frac{33}{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{33}{4 y - 1} d y + \int_{\frac{1}{4}}^{1} \frac{33}{4 y - 1} d y

\int_{0}^{\frac{1}{4}} \frac{33}{4 y - 1} d y =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{\infty} 2 d x

\frac{\partial}{\partial y} (x^{2} cos (x y))

y = x^{3} - 7 x^{2} + 8 x + 1

\int \frac{arctan ( e ^{x} )}{e ^{x}} d x

What is the integral from 0 to 1 of (33)/(4y-1) ?
The integral from 0 to 1 of (33)/(4y-1) is diverges