What is the integral of (3x-1)^{-2} ?

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

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integral of (3x-1)^{-2}

integral

(3 x - 1)^{- 2}

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

Solution steps

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

Apply exponent rule:

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

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

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

Simplify

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

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

Add a constant to the solution

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

x^{4} \cdot n^{- 1} \cdot 1 - x^{2}

y = 2 e^{4} x + 3

- 3 sec (x) \cdot sec^{3} (x)

- sec^{2} (x)

- 3 sec (x) \cdot (sec (x))^{3}

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