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

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Answer 1 · 2016-05-27T04:13:51

$=t-12ln|(t-2)|+17ln|(t-3)|+c$ ,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

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