What is the integral of 28(7x-2)^{-5} ?

The integral of 28(7x-2)^{-5} is -1/((7x-2)^4)+C

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integral of 28(7x-2)^{-5}

\int 28 (7 x - 2)^{- 5} d x

- \frac{1}{( 7 x - 2 ) ^{4}} + C

Solution steps

\int 28 (7 x - 2)^{- 5} d x

Take the constant out:

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

= 28 \cdot \int (7 x - 2)^{- 5} d x

Apply u-substitution

: \int \frac{1}{7 u ^{5}} d u

= 28 \cdot \int \frac{1}{7 u ^{5}} d u

Take the constant out:

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

= 28 \cdot \frac{1}{7} \cdot \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= 28 \cdot \frac{1}{7} (- \frac{1}{4 u ^{4}})

Substitute back

u = 7 x - 2

= 28 \cdot \frac{1}{7} (- \frac{1}{4 ( 7 x - 2 ) ^{4}})

Simplify

28 \cdot \frac{1}{7} (- \frac{1}{4 ( 7 x - 2 ) ^{4}}) : - \frac{1}{( 7 x - 2 ) ^{4}}

= - \frac{1}{( 7 x - 2 ) ^{4}}

Add a constant to the solution

= - \frac{1}{( 7 x - 2 ) ^{4}} + C

\frac{\partial}{\partial x} (\frac{( 2 x - 2 y )}{4 x + 2 y})

\frac{\partial}{\partial x} (t^{2} e^{\frac{x ^{2}}{2 t}})

- x, - 10 x + 14, x = 0

x^{3} e^{2 x}

n = 0 \sum \infty \frac{x ^{n}}{3 \cdot n}

What is the integral of 28(7x-2)^{-5} ?
The integral of 28(7x-2)^{-5} is -1/((7x-2)^4)+C