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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

f (x) = 1 - 2 x

\frac{d}{d x} (e^{- 3 x} sin (3 x))

y = x^{2} + x + 1

\frac{\partial}{\partial y} (x + \frac{1}{2} ln (x^{2} + y^{2}))

\frac{1}{y}

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