What is the integral from 3 to 5 of 1/((x-3)^{3/2)} ?

The integral from 3 to 5 of 1/((x-3)^{3/2)} is diverges

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integral from 3 to 5 of 1/((x-3)^{3/2)}

\int_{3}^{5} \frac{1}{( x - 3 ) ^{\frac{3}{2}}} d x

diverges

Solution steps

\int_{3}^{5} \frac{1}{( x - 3 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int_{0}^{2} \frac{1}{u ^{\frac{3}{2}}} d u

= \int_{0}^{2} \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

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

\int_{- 3}^{1} y^{2} - 3 - (- 2 y) d y

What is the integral from 3 to 5 of 1/((x-3)^{3/2)} ?
The integral from 3 to 5 of 1/((x-3)^{3/2)} is diverges