What is the integral from 0 to 1 of 1/((4x-1)^3) ?

The integral from 0 to 1 of 1/((4x-1)^3) is diverges

Popular >

integral from 0 to 1 of 1/((4x-1)^3)

\int_{0}^{1} \frac{1}{( 4 x - 1 ) ^{3}} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{( 4 x - 1 ) ^{3}} 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{1}{( 4 x - 1 ) ^{3}} d x + \int_{\frac{1}{4}}^{1} \frac{1}{( 4 x - 1 ) ^{3}} d x

\int_{0}^{\frac{1}{4}} \frac{1}{( 4 x - 1 ) ^{3}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{3}^{6} (x + 1) ln (2 x + 4) d x

\int_{0}^{5} e^{0.25 x} d x

\int_{0}^{3} (4 x f (x^{2}) + f (3 x)) d x

\int_{- 4}^{1} - x^{3} - 7 x^{2} - 8 x + 16 d x

\int_{0}^{1} 3 t^{2} e^{t^{3}} d t

What is the integral from 0 to 1 of 1/((4x-1)^3) ?
The integral from 0 to 1 of 1/((4x-1)^3) is diverges