What is the integral from 0 to 1/2 of (2x-1)^{-3} ?

The integral from 0 to 1/2 of (2x-1)^{-3} is diverges

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integral from 0 to 1/2 of (2x-1)^{-3}

\int_{0}^{\frac{1}{2}} (2 x - 1)^{- 3} d x

diverges

Solution steps

\int_{0}^{\frac{1}{2}} (2 x - 1)^{- 3} d x

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\int sin^{2} (a x) d x

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

\frac{d y}{d x} = e^{5 x + 3 y}

y^{^{''}} + 4 y = 3 tan (2 x)

\int sin (2 x - 5) d x

What is the integral from 0 to 1/2 of (2x-1)^{-3} ?
The integral from 0 to 1/2 of (2x-1)^{-3} is diverges