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

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

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

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

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

Solution steps

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

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( 3 + 2 x ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{3}} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 3 + 2 x

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

Simplify

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

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

Add a constant to the solution

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

\int 2 x^{3} ln (x^{2}) d x

y = 3 + 2 x, y = \frac{9 + x}{3}

\frac{d}{d x} (a^{f (x) (x^{2})})

\frac{\partial}{\partial x} (e^{y} (y^{2} - x^{2}))

(tan (x))^{^{'}}

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