What is the integral from 2 to 5 of 1/((x-2)^4) ?

The integral from 2 to 5 of 1/((x-2)^4) is diverges

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integral from 2 to 5 of 1/((x-2)^4)

\int_{2}^{5} \frac{1}{( x - 2 ) ^{4}} d x

diverges

Solution steps

\int_{2}^{5} \frac{1}{( x - 2 ) ^{4}} d x

Apply u-substitution

: \int_{0}^{3} \frac{1}{u ^{4}} d u

= \int_{0}^{3} \frac{1}{u ^{4}} d u

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\int_{0}^{\frac{π}{2}} x cos (7 x) d x

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

\int_{x}^{x^{4}} \frac{y}{x} d y

\int_{0}^{π} sin^{4} (x) cos (x) d x

\int_{0}^{1} e^{x + x^{2}} d x

What is the integral from 2 to 5 of 1/((x-2)^4) ?
The integral from 2 to 5 of 1/((x-2)^4) is diverges