What is the integral from 1 to 5 of 1/((x-4)^2) ?

The integral from 1 to 5 of 1/((x-4)^2) is diverges

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integral from 1 to 5 of 1/((x-4)^2)

\int_{1}^{5} \frac{1}{( x - 4 ) ^{2}} d x

diverges

Solution steps

\int_{1}^{5} \frac{1}{( x - 4 ) ^{2}} d x

Find undefined (singularity) points:

x = 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_{1}^{4} \frac{1}{( x - 4 ) ^{2}} d x + \int_{4}^{5} \frac{1}{( x - 4 ) ^{2}} d x

\int_{1}^{4} \frac{1}{( x - 4 ) ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int - 2 x cos (3 x) d x

\frac{d}{d x} ((x^{2} - 2 x)^{s i n (x)})

y^{^{'}} + 3 y = x + e^{- 2 x}

x \to \infty lim (x - 2^{x})

y^{^{''}} + 8 y^{^{'}} + 9 y = 0

What is the integral from 1 to 5 of 1/((x-4)^2) ?
The integral from 1 to 5 of 1/((x-4)^2) is diverges