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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply u-substitution

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

Substitute back

u = 2 t + 3

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

Simplify

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

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

Add a constant to the solution

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

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