What is the integral of (3x-5)^{-2} ?

The integral of (3x-5)^{-2} is -1/(3(3x-5))+C

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integral of (3x-5)^{-2}

integral

(3 x - 5)^{- 2}

- \frac{1}{3 ( 3 x - 5 )} + C

Solution steps

\int (3 x - 5)^{- 2} d x

Apply exponent rule:

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

= \int \frac{1}{( 3 x - 5 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2}} d u

= \int \frac{1}{3 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{3} (- \frac{1}{u})

Substitute back

u = 3 x - 5

= \frac{1}{3} (- \frac{1}{3 x - 5})

Simplify

\frac{1}{3} (- \frac{1}{3 x - 5}) : - \frac{1}{3 ( 3 x - 5 )}

= - \frac{1}{3 ( 3 x - 5 )}

Add a constant to the solution

= - \frac{1}{3 ( 3 x - 5 )} + C

\frac{x}{y ^{2}}

r^{2} - x^{2}

\frac{x ^{2} + 1}{x}

\frac{2 + e ^{x}}{e ^{x}}

\frac{sin ^{2} ( x )}{e ^{x}}

What is the integral of (3x-5)^{-2} ?
The integral of (3x-5)^{-2} is -1/(3(3x-5))+C