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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = \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 x - 1} d x + \int_{\frac{1}{4}}^{1} \frac{61}{4 x - 1} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (\frac{1}{1 - \frac{2}{x}})

\frac{d}{d t} (cos (8 t))

y^{^{'}} = \frac{1}{( t \cdot y + y + t + 1 )}, y (0) = - 1

f (x) = x + 35, (1, 6)

\frac{d y}{d x} + x^{- 1} y = x y^{2}

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