What is the integral from 3 to 6 of 1/((4-x)^2) ?

The integral from 3 to 6 of 1/((4-x)^2) is diverges

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integral from 3 to 6 of 1/((4-x)^2)

\int_{3}^{6} \frac{1}{( 4 - x ) ^{2}} d x

diverges

Solution steps

\int_{3}^{6} \frac{1}{( 4 - x ) ^{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_{3}^{4} \frac{1}{( 4 - x ) ^{2}} d x + \int_{4}^{6} \frac{1}{( 4 - x ) ^{2}} d x

\int_{3}^{4} \frac{1}{( 4 - x ) ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{s ^{2} + 3}{2 s ^{2} + 5}

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

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2} + 1})

\frac{\partial}{\partial x} (x^{0.5})

What is the integral from 3 to 6 of 1/((4-x)^2) ?
The integral from 3 to 6 of 1/((4-x)^2) is diverges