Medium
Find the function if the antiderivative is H(t)=ln(12t2−12)+CH(t)=\ln (12 t^{2}-12)+CH(t)=ln(12t2−12)+C, where C{C}C is constant.
2tt2−1 \frac{2 t}{t^{2}-1}t2−12t
t2t−1 \frac{t}{2 t-1}2t−1t
4t2−12t3−6t\frac{4 t^{2}-12}{t^{3}-6 t}t3−6t4t2−12
tt+1\frac{t}{t+1} t+1t
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