What is the integral of 2(3x+4)^{-4} ?

The integral of 2(3x+4)^{-4} is -2/(9(3x+4)^3)+C

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integral of 2(3x+4)^{-4}

\int 2 (3 x + 4)^{- 4} d x

- \frac{2}{9 ( 3 x + 4 ) ^{3}} + C

Solution steps

\int 2 (3 x + 4)^{- 4} d x

Take the constant out:

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

= 2 \cdot \int (3 x + 4)^{- 4} d x

Apply u-substitution

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

= 2 \cdot \int \frac{1}{3 u ^{4}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= 2 \cdot \frac{1}{3} (- \frac{1}{3 u ^{3}})

Substitute back

u = 3 x + 4

= 2 \cdot \frac{1}{3} (- \frac{1}{3 ( 3 x + 4 ) ^{3}})

Simplify

2 \cdot \frac{1}{3} (- \frac{1}{3 ( 3 x + 4 ) ^{3}}) : - \frac{2}{9 ( 3 x + 4 ) ^{3}}

= - \frac{2}{9 ( 3 x + 4 ) ^{3}}

Add a constant to the solution

= - \frac{2}{9 ( 3 x + 4 ) ^{3}} + C

f (x) = \frac{x}{x + 1}, (1, 2)

\frac{d}{d x} (\frac{x}{1 + sin ( x )})

4^{3 - x^{2}}

\frac{16}{s ^{2} + 16}

4 x^{^{''}} + 2 x^{^{'}} + 4 x = 100

What is the integral of 2(3x+4)^{-4} ?
The integral of 2(3x+4)^{-4} is -2/(9(3x+4)^3)+C