What is the integral of-1120(1+4t)^{-3} ?

The integral of-1120(1+4t)^{-3} is (140)/((1+4t)^2)+C

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integral of-1120(1+4t)^{-3}

\int - 1120 (1 + 4 t)^{- 3} d t

\frac{140}{( 1 + 4 t ) ^{2}} + C

Solution steps

\int - 1120 (1 + 4 t)^{- 3} d t

Take the constant out:

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

= - 1120 \cdot \int (1 + 4 t)^{- 3} d t

Apply u-substitution

: \int \frac{1}{4 u ^{3}} d u

= - 1120 \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

= - 1120 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= - 1120 \cdot \frac{1}{4} (- \frac{1}{2 u ^{2}})

Substitute back

u = 1 + 4 t

= - 1120 \cdot \frac{1}{4} (- \frac{1}{2 ( 1 + 4 t ) ^{2}})

Simplify

- 1120 \cdot \frac{1}{4} (- \frac{1}{2 ( 1 + 4 t ) ^{2}}) : \frac{140}{( 1 + 4 t ) ^{2}}

= \frac{140}{( 1 + 4 t ) ^{2}}

Add a constant to the solution

= \frac{140}{( 1 + 4 t ) ^{2}} + C

f (x) = 5 x^{2} - x^{3}

\int_{\frac{π}{2}}^{π} 3 x - sin (x) d x

\int \frac{x}{( x ^{2} - 1 ) [ ln ( x ^{2} - 1 ) ] ^{2}} d x

y^{^{'}} = 5 y^{2} sin (x)

\frac{d}{d x} (x^{c o t (x)})

What is the integral of-1120(1+4t)^{-3} ?
The integral of-1120(1+4t)^{-3} is (140)/((1+4t)^2)+C