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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 2 t + 3

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

Simplify

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

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

Add a constant to the solution

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

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