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

The integral of 1/((x-4)^2) is -1/(x-4)+C

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integral of 1/((x-4)^2)

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

- \frac{1}{x - 4} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x - 4

= - \frac{1}{x - 4}

Add a constant to the solution

= - \frac{1}{x - 4} + C

\frac{d}{d x} (sin (2 x) + cos^{2} (x))

\frac{d}{d x} (4 ln (5 x))

\frac{d}{d x} ((x - 4) e^{3 x})

e^{- 2 x}

x \to 0 lim (coth (x))