What is the integral from 0 to 1 of (42)/(4y-1) ?

The integral from 0 to 1 of (42)/(4y-1) is diverges

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integral from 0 to 1 of (42)/(4y-1)

\int_{0}^{1} \frac{42}{4 y - 1} d y

diverges

Solution steps

\int_{0}^{1} \frac{42}{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{42}{4 y - 1} d y + \int_{\frac{1}{4}}^{1} \frac{42}{4 y - 1} d y

\int_{0}^{\frac{1}{4}} \frac{42}{4 y - 1} d y =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 2 \sum \infty n \cdot (\frac{3}{4})^{n}

y^{^{''}} = 6 x + 2, y (1) = - 1, y^{^{'}} (1) = 7

\frac{\partial}{\partial z} (arctan (x^{2} y z^{2}))

\frac{\partial}{\partial z} (x arctan (yz))

\frac{\partial}{\partial x} ((2 x + 3 y)^{2})

What is the integral from 0 to 1 of (42)/(4y-1) ?
The integral from 0 to 1 of (42)/(4y-1) is diverges