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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{25} 5 x d x

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\int_{- 1}^{1} (x^{2} + 2 x + 1) d x

\int_{0}^{\frac{π}{2}} 4 cos^{4} (x) d x

\int_{2}^{4} x - 2 d x

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