What is the integral from 0 to 4 of 1/((x-4)^2) ?

The integral from 0 to 4 of 1/((x-4)^2) is diverges

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integral from 0 to 4 of 1/((x-4)^2)

\int_{0}^{4} \frac{1}{( x - 4 ) ^{2}} d x

diverges

Solution steps

\int_{0}^{4} \frac{1}{( x - 4 ) ^{2}} d x

Apply u-substitution

: \int_{- 4}^{0} \frac{1}{u ^{2}} d u

= \int_{- 4}^{0} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{- 4}^{0}

= [- \frac{1}{u}]_{- 4}^{0}

Compute the boundaries:

diverges

= diverges

\int \frac{4 - ln ( x + 3 )}{x + 3} d x

\frac{d}{d y} (4 x y)

x \to - 2 lim (- 4 x^{3} + 4 x + 5)

\frac{1}{x}, \frac{1}{x ^{2}}, x = 2

\frac{2}{3 s + 5}

What is the integral from 0 to 4 of 1/((x-4)^2) ?
The integral from 0 to 4 of 1/((x-4)^2) is diverges