I just don't understand how my teacher has got from part 1). to 2). Where has the 1/5 come from?

Q(t) = ∫(1/((t^2)−t−6))dt

 =(1/5)∫(1/(t−3) - 1/(t+2))dt
 =(1/5)ln(t−3) − ln(t+2)
  • $\begingroup$ This question can better be posted on mathematics.stackexchange.com $\endgroup$
    – dreamer
    Apr 28, 2015 at 9:44

1 Answer 1


$(t^2-t-6)^{-1}=((t-3)(t+2))^{-1}$ but it is not directly integrable.

So, your teacher added $(t+2)-(t-3)=5$ in the numerator. Hence, divide by $5$.

Now you can easily integrate $\frac{1}{5}\times\frac{(t+2)-(t-3)}{(t-3)(t+2)}=\frac{1}{5}\times\Big(\frac{1}{t-3}-\frac{1}{t+2}\Big)$


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