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

May 27, 2016

$= t - 12 \ln | \left(t - 2\right) | + 17 \ln | \left(t - 3\right) | + c$ ,where c = integration constant

Explanation:

I$= \int \frac{{t}^{2} + 8}{{t}^{2} - 5 t + 6} \mathrm{dt}$

$= \int \frac{\left({t}^{2} - 5 t + 6\right) + \left(5 t + 2\right)}{{t}^{2} - 5 t + 6} \mathrm{dt}$

$= \int \frac{{t}^{2} - 5 t + 6}{{t}^{2} - 5 t + 6} \mathrm{dt} + \int \frac{5 t + 2}{\text{(t-3)(t-2)}} \mathrm{dt}$

Now
Let $a \times \left(t - 3\right) + b \times \left(t - 2\right) = 5 t + 2$

for t= 3 ,
$a \times \left(3 - 3\right) + b \times \left(3 - 2\right) = 5 \times 3 + 2 \implies b = 17$

for t= 2 ,
$a \times \left(2 - 3\right) + b \times \left(2 - 2\right) = 5 \times 2 + 2 \implies a = - 12$

I$= \int \mathrm{dt} + \int \frac{- 12 \times \left(t - 3\right) + 17 \times \left(t - 2\right)}{\text{(t-3)(t-2)}} \mathrm{dt}$

$= \int \mathrm{dt} - 12 \int \frac{\mathrm{dt}}{t - 2} + 17 \int \frac{\mathrm{dt}}{t - 3}$

$= t - 12 \ln | \left(t - 2\right) | + 17 \ln | \left(t - 3\right) | + c$ ,where c = integration constant