What is the integral of 1/((x-6)^2) ?

The integral of 1/((x-6)^2) is -1/(x-6)+C

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integral of 1/((x-6)^2)

\int \frac{1}{( x - 6 ) ^{2}} d x

- \frac{1}{x - 6} + C

Solution steps

\int \frac{1}{( x - 6 ) ^{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 - 6

= - \frac{1}{x - 6}

Add a constant to the solution

= - \frac{1}{x - 6} + C

\frac{\partial}{\partial x} (sin (x^{2}) + x y^{3})

f (x) = \frac{1 + csc ( x ^{3} )}{1 - cot ( x ^{3} )}

\int (4 - x)^{3} d x

n = 1 \sum \infty \frac{n ^{2}}{n + 1}

y = - \frac{x}{9 y}