What is the integral from 0 to 3 of 1/((x-3)^3) ?

The integral from 0 to 3 of 1/((x-3)^3) is diverges

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integral from 0 to 3 of 1/((x-3)^3)

\int_{0}^{3} \frac{1}{( x - 3 ) ^{3}} d x

diverges

Solution steps

\int_{0}^{3} \frac{1}{( x - 3 ) ^{3}} d x

Apply u-substitution

: \int_{- 3}^{0} \frac{1}{u ^{3}} d u

= \int_{- 3}^{0} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{- 3}^{0}

= [- \frac{1}{2 u ^{2}}]_{- 3}^{0}

Compute the boundaries:

diverges

= diverges

x \to 0 lim (5 + x^{2})

\int 4 x cos (7 x) d x

f (x) = \frac{8 t}{8 + t}

\frac{- 2 s - 5}{7 s ^{2} + 26 s + 25}

\frac{d}{d x} ((9 x^{6} + 2 x^{3})^{4})

What is the integral from 0 to 3 of 1/((x-3)^3) ?
The integral from 0 to 3 of 1/((x-3)^3) is diverges