evaluate $int\frac{3}{(t^{2}-9)^{2}}dt$.

Explanation:

Step1: Use partial - fraction decomposition

First, factor $t^{2}-9=(t - 3)(t + 3)$. Then, $\frac{3}{(t^{2}-9)^{2}}=\frac{3}{(t - 3)^{2}(t + 3)^{2}}=\frac{A}{t - 3}+\frac{B}{(t - 3)^{2}}+\frac{C}{t+3}+\frac{D}{(t + 3)^{2}}$. Cross - multiply: $3=A(t - 3)(t + 3)^{2}+B(t + 3)^{2}+C(t + 3)(t - 3)^{2}+D(t - 3)^{2}$. Let $t = 3$, then $3=B(3 + 3)^{2}$, so $B=\frac{3}{36}=\frac{1}{12}$. Let $t=-3$, then $3=D(-3 - 3)^{2}$, so $D=\frac{3}{36}=\frac{1}{12}$. Expand the right - hand side: [ \begin{align*} A(t - 3)(t + 3)^{2}+B(t + 3)^{2}+C(t + 3)(t - 3)^{2}+D(t - 3)^{2}&=A(t - 3)(t^{2}+6t + 9)+B(t^{2}+6t + 9)+C(t + 3)(t^{2}-6t + 9)+D(t^{2}-6t + 9)\ &=A(t^{3}+6t^{2}+9t-3t^{2}-18t - 27)+B(t^{2}+6t + 9)+C(t^{3}-6t^{2}+9t+3t^{2}-18t + 27)+D(t^{2}-6t + 9)\ &=A(t^{3}+3t^{2}-9t - 27)+B(t^{2}+6t + 9)+C(t^{3}-3t^{2}-9t + 27)+D(t^{2}-6t + 9) \end{align*} ] Collect the coefficients of $t^{3}$: $0=A + C$. Collect the coefficients of $t^{2}$: $0 = 3A + B-3C+D$. Since $B = D=\frac{1}{12}$, we have $0 = 3A+\frac{1}{12}-3C+\frac{1}{12}$, and since $C=-A$, $0 = 3A+\frac{1}{6}+3A$, so $6A=-\frac{1}{6}$, $A =-\frac{1}{36}$ and $C=\frac{1}{36}$. So $\frac{3}{(t^{2}-9)^{2}}=-\frac{1}{36(t - 3)}+\frac{1}{12(t - 3)^{2}}+\frac{1}{36(t + 3)}+\frac{1}{12(t + 3)^{2}}$.

Step2: Integrate term - by - term

[ \begin{align*} \int\frac{3}{(t^{2}-9)^{2}}dt&=-\frac{1}{36}\int\frac{1}{t - 3}dt+\frac{1}{12}\int\frac{1}{(t - 3)^{2}}dt+\frac{1}{36}\int\frac{1}{t + 3}dt+\frac{1}{12}\int\frac{1}{(t + 3)^{2}}dt\ &=-\frac{1}{36}\ln|t - 3|-\frac{1}{12(t - 3)}+\frac{1}{36}\ln|t + 3|-\frac{1}{12(t + 3)}+C\ &=\frac{1}{36}\ln\left|\frac{t + 3}{t - 3}\right|-\frac{t}{6(t^{2}-9)}+C \end{align*} ]

Answer:

$\frac{1}{36}\ln\left|\frac{t + 3}{t - 3}\right|-\frac{t}{6(t^{2}-9)}+C$