What is the integral of-1280(1+4x)^{-3} ?

The integral of-1280(1+4x)^{-3} is (160)/((1+4x)^2)+C

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integral of-1280(1+4x)^{-3}

\int - 1280 (1 + 4 x)^{- 3} d x

\frac{160}{( 1 + 4 x ) ^{2}} + C

Solution steps

\int - 1280 (1 + 4 x)^{- 3} d x

Take the constant out:

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

= - 1280 \cdot \int (1 + 4 x)^{- 3} d x

Apply u-substitution

= - 1280 \cdot \int \frac{1}{4 u ^{3}} d u

Take the constant out:

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

= - 1280 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

= - 1280 \cdot \frac{1}{4} (- \frac{1}{2 u ^{2}})

Substitute back

u = 1 + 4 x

= - 1280 \cdot \frac{1}{4} (- \frac{1}{2 ( 1 + 4 x ) ^{2}})

Simplify

- 1280 \cdot \frac{1}{4} (- \frac{1}{2 ( 1 + 4 x ) ^{2}}) : \frac{160}{( 1 + 4 x ) ^{2}}

= \frac{160}{( 1 + 4 x ) ^{2}}

Add a constant to the solution

= \frac{160}{( 1 + 4 x ) ^{2}} + C

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What is the integral of-1280(1+4x)^{-3} ?
The integral of-1280(1+4x)^{-3} is (160)/((1+4x)^2)+C