What is the integral of (x-3)^{-2} ?

The integral of (x-3)^{-2} is -1/(x-3)+C

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integral of (x-3)^{-2}

\int (x - 3)^{- 2} d x

- \frac{1}{x - 3} + C

Solution steps

\int (x - 3)^{- 2} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( x - 3 ) ^{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 - 3

= - \frac{1}{x - 3}

Add a constant to the solution

= - \frac{1}{x - 3} + C

x \to 0 lim (\frac{2 ^{x} - 3 ^{x}}{x})

y = (x^{3} - 16 x)^{12}, (- 4, 0)

\frac{5 x ^{2}}{x}

y^{^{'}} + y = e^{- x}

\frac{8 - sec ( x )}{tan ( x )}