# Thread: Integration using partial fractions

1. ## Integration using partial fractions

S 1/[(t+4)(t-1)]dt is my problem. I'm having trouble understanding how to use the '1' here.

Thanks!

2. Originally Posted by JYZero
S 1/[(t+4)(t-1)]dt is my problem. I'm having trouble understanding how to use the '1' here.

Thanks!
Let $\frac{A}{t + 4} + \frac{B}{t - 1} = \frac{1}{(t + 4)(t - 1)}$

$\frac{A(t - 1) + B(t + 4)}{(t + 4)(t - 1)} = \frac{1}{(t + 4)(t - 1)}$

$A(t - 1) + B(t + 4) = 1$

$At - A + Bt + 4B = 1$

$(A + B)t + 4B - A = 0t + 1$.

So $A + B = 0$ and $4B - A = 1$

$4B - 1 = A$

So $4B - 1 + B = 0$

$5B - 1 = 0$

$B = \frac{1}{5}$

$A + B = 0$

$A + \frac{1}{5} = 0$

$A = -\frac{1}{5}$.

So $\int{\frac{1}{(t + 4)(t - 1)}\,dt} = \int{-\frac{1}{5}\left(\frac{1}{t + 4}\right) + \frac{1}{5}\left(\frac{1}{t - 1}\right)\,dt}$

$= \frac{1}{5}\int{\frac{1}{t - 1} - \frac{1}{t + 4}\,dt}$

$= \frac{1}{5}\left[\ln{|t - 1|} - \ln{|t + 4|}\right] + C$

$= \frac{1}{5}\ln{\left|\frac{t - 1}{t + 4}\right|} + C$.

3. Originally Posted by JYZero
S 1/[(t+4)(t-1)]dt is my problem. I'm having trouble understanding how to use the '1' here.
$\frac{1}{(t+4)(t-1)}=\frac{1}{5}\cdot \frac{(t+4)-(t-1)}{(t+4)(t-1)}=\frac{1}{5}\left( \frac{1}{t-1}-\frac{1}{t+4} \right).$