How do you integrate int (6(5 - x))/( (x - 7)(4 - x)) using partial fractions?

Apr 13, 2017

$\int \frac{6 \left(5 - x\right)}{\left(x - 7\right) \left(4 - x\right)} \mathrm{dx} = 4 \ln | x - 7 | + 2 \ln | x - 4 | + C$

Explanation:

By multiplying the numerator and the denominator by $- 1$,

$\frac{6 \left(x - 5\right)}{\left(x - 7\right) \left(x - 4\right)} = \frac{A}{x - 7} + \frac{B}{x - 4} = \frac{A \left(x - 4\right) + B \left(x - 7\right)}{\left(x - 7\right) \left(x - 4\right)}$

By matching the numerators,

$A \left(x - 4\right) + B \left(x - 7\right) = 6 \left(x - 5\right)$

To find $A$, set $x = 7 R i g h t a r r o w 3 A = 12 R i g h t a r r o w A = 4$

To find $B$, set $x = 4 R i g h t a r r o w - 3 B = - 6 R i g h t a r r o w B = 2$

So, we have

int(6(5-x))/((x-7)(4-x))dx =int(4/(x-7)+2/(x-4) )dx

By Log Rule,

$= 4 \ln | x - 7 | + 2 \ln | x - 4 | + C$

I hope that this was clear.