What is the integral from 0 to 1 of (61)/(4y-1) ?

The integral from 0 to 1 of (61)/(4y-1) is diverges

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integral from 0 to 1 of (61)/(4y-1)

\int_{0}^{1} \frac{61}{4 y - 1} d y

diverges

Solution steps

\int_{0}^{1} \frac{61}{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{61}{4 y - 1} d y + \int_{\frac{1}{4}}^{1} \frac{61}{4 y - 1} d y

\int_{0}^{\frac{1}{4}} \frac{61}{4 y - 1} d y =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d y}{d x}, ln (x^{2} - 24 y) = x - y - 4

f (x) = - 2 - 6 x^{2}, (5, - 152)

\int sin (x) cos^{2} (x) d x

\frac{3 s - 4}{s ^{2} + 4 s + 8}

\frac{d}{d y} (\frac{y - 6}{y ^{2} - 2 y + 12})

What is the integral from 0 to 1 of (61)/(4y-1) ?
The integral from 0 to 1 of (61)/(4y-1) is diverges