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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 3

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d y}{d x}, y = 2 x^{3}

\int e^{3 x} sin (4 x) d x

f (x) = \frac{3}{x ^{2}} - \frac{1}{x}

1 + 0.1 e^{- 0.02 t}

x^{2} y^{^{''}} - 2 x y^{^{'}} + 2 y = 4 x^{2}

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