What is the integral from 2 to 4 of 1/((x-2)^3) ?

The integral from 2 to 4 of 1/((x-2)^3) is diverges

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integral from 2 to 4 of 1/((x-2)^3)

\int_{2}^{4} \frac{1}{( x - 2 ) ^{3}} d x

diverges

Solution steps

\int_{2}^{4} \frac{1}{( x - 2 ) ^{3}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\int \frac{1}{5 ( x ^{2} + 1 )} d x

\int \frac{9 + 9 x}{1 + x ^{2}} d x

\frac{d y}{d x} = \frac{x ^{2} + 3 y ^{2}}{2 x y}

f (x) = \frac{x ^{2}}{x ^{2} + 1}

x \to 0 + lim ((1 + 2 x)^{\frac{3}{x}})

What is the integral from 2 to 4 of 1/((x-2)^3) ?
The integral from 2 to 4 of 1/((x-2)^3) is diverges