What is the integral from 6 to 7 of 1/((x-6)^{5/4)} ?

The integral from 6 to 7 of 1/((x-6)^{5/4)} is diverges

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integral from 6 to 7 of 1/((x-6)^{5/4)}

\int_{6}^{7} \frac{1}{( x - 6 ) ^{\frac{5}{4}}} d x

diverges

Solution steps

\int_{6}^{7} \frac{1}{( x - 6 ) ^{\frac{5}{4}}} d x

Apply u-substitution

: \int_{0}^{1} \frac{1}{u ^{\frac{5}{4}}} d u

= \int_{0}^{1} \frac{1}{u ^{\frac{5}{4}}} d u

Apply the Power Rule

: [- \frac{4}{u ^{\frac{1}{4}}}]_{0}^{1}

= [- \frac{4}{u ^{\frac{1}{4}}}]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int_{0}^{\frac{π}{2}} x cos (9 x) d x

\int_{0}^{1} \frac{1}{x ^{π}} d x

\int_{0}^{1} \frac{( 1 - x ) ^{2}}{4} d x

\int_{0}^{π} (x + 1) d x

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

What is the integral from 6 to 7 of 1/((x-6)^{5/4)} ?
The integral from 6 to 7 of 1/((x-6)^{5/4)} is diverges