What is the integral from 0 to 8 of 4/((x-6)^3) ?

The integral from 0 to 8 of 4/((x-6)^3) is diverges

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integral from 0 to 8 of 4/((x-6)^3)

\int_{0}^{8} \frac{4}{( x - 6 ) ^{3}} d x

diverges

Solution steps

\int_{0}^{8} \frac{4}{( x - 6 ) ^{3}} d x

Find undefined (singularity) points:

x = 6

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}^{6} \frac{4}{( x - 6 ) ^{3}} d x + \int_{6}^{8} \frac{4}{( x - 6 ) ^{3}} d x

\int_{0}^{6} \frac{4}{( x - 6 ) ^{3}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

What is the integral from 0 to 8 of 4/((x-6)^3) ?
The integral from 0 to 8 of 4/((x-6)^3) is diverges