match each function to its antiderivative, where t > 3 and c is a constant. antiderivative function…

Explanation:

Step1: Recall the chain - rule for derivatives of logarithmic functions

The derivative of $\ln(u)$ with respect to $t$ is $\frac{u'}{u}$, where $u$ is a function of $t$.

Step2: For $H(t)=\ln(12t^{2}-12)+C$

Let $u = 12t^{2}-12$, then $u'=24t$. So $\frac{dH}{dt}=\frac{24t}{12t^{2}-12}=\frac{2t}{t^{2}-1}$.

Step3: For $H(t)=\ln(t^{4}-6t^{2})+C$

Let $u=t^{4}-6t^{2}$, then $u' = 4t^{3}-12t$. So $\frac{dH}{dt}=\frac{4t^{3}-12t}{t^{4}-6t^{2}}=\frac{4t^{2}-12}{t^{3}-6t}$.

Step4: For $H(t)=\ln(4t^{3}-12t)+C$

Let $u = 4t^{3}-12t$, then $u'=12t^{2}-12$. So $\frac{dH}{dt}=\frac{12t^{2}-12}{4t^{3}-12t}=\frac{3t^{2}-3}{t^{3}-3t}$.

Answer:

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