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

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

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

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

\frac{3}{t + 2} + C

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= - 3 (- \frac{1}{u})

Substitute back

u = t + 2

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

Simplify

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

= \frac{3}{t + 2}

Add a constant to the solution

= \frac{3}{t + 2} + C

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