What is the integral of (5x-4)^{-3} ?

The integral of (5x-4)^{-3} is -1/(10(5x-4)^2)+C

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integral of (5x-4)^{-3}

integral

(5 x - 4)^{- 3}

- \frac{1}{10 ( 5 x - 4 ) ^{2}} + C

Solution steps

\int (5 x - 4)^{- 3} d x

Apply exponent rule:

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

= \int \frac{1}{( 5 x - 4 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{5 u ^{3}} d u

= \int \frac{1}{5 u ^{3}} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= \frac{1}{5} (- \frac{1}{2 u ^{2}})

Substitute back

u = 5 x - 4

= \frac{1}{5} (- \frac{1}{2 ( 5 x - 4 ) ^{2}})

Simplify

\frac{1}{5} (- \frac{1}{2 ( 5 x - 4 ) ^{2}}) : - \frac{1}{10 ( 5 x - 4 ) ^{2}}

= - \frac{1}{10 ( 5 x - 4 ) ^{2}}

Add a constant to the solution

= - \frac{1}{10 ( 5 x - 4 ) ^{2}} + C

∣ x ∣^{2}

sin (y^{3})

\frac{( e ^{7} x )}{( 2 + e ^{7} x )}

x^{2} - 8 x + 7

\frac{x}{( 1 + x ) ^{3}}

What is the integral of (5x-4)^{-3} ?
The integral of (5x-4)^{-3} is -1/(10(5x-4)^2)+C