Question

match each function to its antiderivative, where ( t > 3 ) and ( c ) is a constant.
antiderivative | function
( h(t) = ln(12t^2 - 12) + c ) | ( h(t) = \frac{4t^2 - 12}{t^3 - 6t} )
( h(t) = ln(t^4 - 6t^2) + c ) | ( h(t) = \frac{3t^2 - 3}{t^3 - 3t} )
( h(t) = ln(4t^3 - 12t) + c ) | ( h(t) = \frac{2t}{t^2 - 1} )

Explanation:

Step1: Differentiate $H(t)=\ln(12t^2-12)+C$

Use chain rule: $\frac{d}{dt}\ln(u)=\frac{u'}{u}$. Let $u=12t^2-12$, $u'=24t$.
$H'(t)=\frac{24t}{12t^2-12}=\frac{2t}{t^2-1}$

Step2: Differentiate $H(t)=\ln(t^4-6t^2)+C$

Let $u=t^4-6t^2$, $u'=4t^3-12t$.
$H'(t)=\frac{4t^3-12t}{t^4-6t^2}=\frac{4t(t^2-3)}{t^2(t^2-6)}=\frac{4t^2-12}{t^3-6t}$

Step3: Differentiate $H(t)=\ln(4t^3-12t)+C$

Let $u=4t^3-12t$, $u'=12t^2-12$.
$H'(t)=\frac{12t^2-12}{4t^3-12t}=\frac{3t^2-3}{t^3-3t}$

Answer:

1. $H(t)=\ln(12t^2-12)+C$ matches $h(t)=\frac{2t}{t^2-1}$
2. $H(t)=\ln(t^4-6t^2)+C$ matches $h(t)=\frac{4t^2-12}{t^3-6t}$
3. $H(t)=\ln(4t^3-12t)+C$ matches $h(t)=\frac{3t^2-3}{t^3-3t}$