How do you integrate #int (t^2 + 8) / (t^2 - 5t + 6)# using partial fractions?

Question

Christopher Agresta · Accepted Answer

#=t-12ln|(t-2)|+17ln|(t-3)|+c# ,where c = integration constant

I#=int (t^2 + 8) / (t^2 - 5t + 6)dt# #=int ((t^2 -5t+ 6)+(5t+2)) / (t^2 - 5t + 6)dt# #=int (t^2 -5t+ 6)/ (t^2 - 5t + 6)dt+int(5t+2) / "(t-3)(t-2)"dt# Now Let #axx(t-3)+bxx(t-2)=5t+2# for t= 3 , #axx(3-3)+bxx(3-2)=5xx3+2=>b=17# for t= 2 , #axx(2-3)+bxx(2-2)=5xx2+2=>a=-12# I#=intdt+int(-12xx(t-3)+17xx(t-2)) / "(t-3)(t-2)"dt# #=intdt-12int(dt)/(t-2)+17int(dt)/(t-3)# #=t-12ln|(t-2)|+17ln|(t-3)|+c# ,where c = integration constant

Isabella Ackerman · Answer

undefined To integrate $\frac{t^2 + 8}{t^2 - 5t + 6}$ using partial fractions, follow these steps:

1. Factor the denominator $t^2 - 5t + 6$. It factors into $(t - 2)(t - 3)$.
2. Decompose the rational function into partial fractions based on its factorization. The form of the partial fractions will be:
   $$\frac{t^2 + 8}{(t - 2)(t - 3)} = \frac{A}{t - 2} + \frac{B}{t - 3}$$
3. Multiply both sides of the equation by the denominator $(t - 2)(t - 3)$ to clear the fractions.
4. Expand and combine like terms to solve for $A$ and $B$.
5. Once you've found the values of $A$ and $B$, rewrite the original integral with the decomposed fractions.
6. Integrate each term separately.
7. Finally, combine the results to obtain the overall integral.

The steps may seem a bit involved, but they follow a straightforward process to decompose the fraction into simpler terms that are easier to integrate.