What is the integral from 6 to 7 of 1/((x-6)^{3/2)} ?

The integral from 6 to 7 of 1/((x-6)^{3/2)} is diverges

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integral from 6 to 7 of 1/((x-6)^{3/2)}

\int_{6}^{7} \frac{1}{( x - 6 ) ^{\frac{3}{2}}} d x

diverges

Solution steps

\int_{6}^{7} \frac{1}{( x - 6 ) ^{\frac{3}{2}}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

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What is the integral from 6 to 7 of 1/((x-6)^{3/2)} ?
The integral from 6 to 7 of 1/((x-6)^{3/2)} is diverges