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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

(\frac{3}{5}, \frac{2}{3}), (- \frac{3}{5}, \frac{3}{4})

\int (e^{t})^{2} d t

\int x^{3} sec^{2} (y) d y

y^{^{''}} + \frac{y ^{^{'}}}{x} = 2 x + 1

arctan (x)

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