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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

\int 7 v (v^{2} + 4)^{2} d v

\int 2 + 2 sin (x) d x

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

f (x) = \frac{- 8 x}{x ^{2} + 1}, at x = 1

\frac{\partial}{\partial x} (2 x^{2} y + 4)

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