How do you integrate int (t^2 + 8) / (t^2 - 5t + 6)∫t2+8t2−5t+6 using partial fractions?

1 Answer
P dilip_k
May 27, 2016

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

Explanation:

I=int (t^2 + 8) / (t^2 - 5t + 6)dt=∫t2+8t2−5t+6dt

=int ((t^2 -5t+ 6)+(5t+2)) / (t^2 - 5t + 6)dt=∫(t2−5t+6)+(5t+2)t2−5t+6dt

=int (t^2 -5t+ 6)/ (t^2 - 5t + 6)dt+int(5t+2) / "(t-3)(t-2)"dt=∫t2−5t+6t2−5t+6dt+∫5t+2(t-3)(t-2)dt

Now
Let axx(t-3)+bxx(t-2)=5t+2a×(t−3)+b×(t−2)=5t+2

for t= 3 ,
axx(3-3)+bxx(3-2)=5xx3+2=>b=17a×(3−3)+b×(3−2)=5×3+2⇒b=17

for t= 2 ,
axx(2-3)+bxx(2-2)=5xx2+2=>a=-12a×(2−3)+b×(2−2)=5×2+2⇒a=−12

I=intdt+int(-12xx(t-3)+17xx(t-2)) / "(t-3)(t-2)"dt=∫dt+∫−12×(t−3)+17×(t−2)(t-3)(t-2)dt

=intdt-12int(dt)/(t-2)+17int(dt)/(t-3)=∫dt−12∫dtt−2+17∫dtt−3

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

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