What is the integral of-64(1+4t)^{-3} ?

The integral of-64(1+4t)^{-3} is 8/((1+4t)^2)+C

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integral of-64(1+4t)^{-3}

\int - 64 (1 + 4 t)^{- 3} d t

\frac{8}{( 1 + 4 t ) ^{2}} + C

Solution steps

\int - 64 (1 + 4 t)^{- 3} d t

Take the constant out:

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

= - 64 \cdot \int (1 + 4 t)^{- 3} d t

Apply u-substitution

= - 64 \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

= - 64 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

= - 64 \cdot \frac{1}{4} (- \frac{1}{2 u ^{2}})

Substitute back

u = 1 + 4 t

= - 64 \cdot \frac{1}{4} (- \frac{1}{2 ( 1 + 4 t ) ^{2}})

Simplify

- 64 \cdot \frac{1}{4} (- \frac{1}{2 ( 1 + 4 t ) ^{2}}) : \frac{8}{( 1 + 4 t ) ^{2}}

= \frac{8}{( 1 + 4 t ) ^{2}}

Add a constant to the solution

= \frac{8}{( 1 + 4 t ) ^{2}} + C

What is the integral of-64(1+4t)^{-3} ?
The integral of-64(1+4t)^{-3} is 8/((1+4t)^2)+C