What is the integral from 1 to 4 of 1/((x-1)^3) ?

The integral from 1 to 4 of 1/((x-1)^3) is diverges

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integral from 1 to 4 of 1/((x-1)^3)

\int_{1}^{4} \frac{1}{( x - 1 ) ^{3}} d x

diverges

Solution steps

\int_{1}^{4} \frac{1}{( x - 1 ) ^{3}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\int_{0}^{\frac{π}{4}} 1 + sin^{2} (x) d x

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

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

\int_{1}^{9} x^{3} d x

\int_{0.3}^{0.6} x d x

What is the integral from 1 to 4 of 1/((x-1)^3) ?
The integral from 1 to 4 of 1/((x-1)^3) is diverges