What is the integral of (9x-2)^{-3} ?

The integral of (9x-2)^{-3} is -1/(18(9x-2)^2)+C

integral of (9x-2)^{-3}

\int (9 x - 2)^{- 3} d x

- \frac{1}{18 ( 9 x - 2 ) ^{2}} + C

Solution steps

\int (9 x - 2)^{- 3} d x

Apply exponent rule:

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

= \int \frac{1}{( 9 x - 2 ) ^{3}} d x

Apply u-substitution

= \int \frac{1}{9 u ^{3}} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

= \frac{1}{9} (- \frac{1}{2 u ^{2}})

Substitute back

u = 9 x - 2

= \frac{1}{9} (- \frac{1}{2 ( 9 x - 2 ) ^{2}})

Simplify

\frac{1}{9} (- \frac{1}{2 ( 9 x - 2 ) ^{2}}) : - \frac{1}{18 ( 9 x - 2 ) ^{2}}

= - \frac{1}{18 ( 9 x - 2 ) ^{2}}

Add a constant to the solution

= - \frac{1}{18 ( 9 x - 2 ) ^{2}} + C

\int 5 r e^{\frac{r}{2}} d r

d er i v a t i v e f (x) = x^{2} ln (8 - 5 x^{2})

a re a y = x^{3} - 10 x, y = 6 x

\frac{d y}{d x} = x^{6} y^{- 5}

\int_{- 5}^{5} \frac{1}{∣ x ∣ ^{\frac{2}{3}}} d x

What is the integral of (9x-2)^{-3} ?
The integral of (9x-2)^{-3} is -1/(18(9x-2)^2)+C