How do you integrate 7/(x-3)^2 using partial fractions?

Oct 12, 2017

$\frac{- 7}{x - 3} + c$

Explanation:

$\frac{7}{x - 3} ^ 2 = \left(\frac{A 1}{x - 3}\right) + \left(\frac{A 2}{x - 3} ^ 2\right)$
$A 1 \cdot \left(x - 3\right) + A 2 = 7$
For x = 3,
$A 2 = 7$
Comparing coefficient of x, we get $A = 0$
Let $t = \left(x - 3\right) , \mathrm{dt} = \mathrm{dx}$
$\int \left(\frac{7}{t} ^ 2\right) \mathrm{dt} = \left(- \frac{7}{t}\right) + c$
Substituting value of $t = \left(x - 3\right)$
$= \frac{- 7}{x - 3} + c$