How do you integrate $\int \frac{x}{(x - 6) (x - 3) d x}$ $int x/((x-6)(x-3) dx$ using partial fractions?

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Answer 1 · 2015-11-24T21:07:28

$\int \frac{x}{(x - 6) (x - 3)} d x = 2 ln | x - 6 | - ln | x - 3 | + C$ $intx/((x-6)(x-3))dx = 2ln|x-6| - ln|x-3| + C$

To find the partial fraction decomposition, we begin with

$\frac{x}{(x - 6) (x - 3)} = \frac{A}{x - 6} + \frac{B}{x - 3}$ $x/((x-6)(x-3)) = A/(x-6) + B/(x-3)$

Multiplying by $(x - 6) (x - 3)$ $(x-6)(x-3)$ gives us

$x = A (x - 3) + B (x - 6) = (A + B) x + (- 3 A - 6 B)$ $x = A(x-3)+B(x-6) = (A+B)x + (-3A - 6B)$

Matching respective coefficients on each side gives us the system of equations

${(A + B = 1), (-3A -6B = 0):}$

Now all we need to do is solve for $A$ and $B$ . For example, using elimination,

$(-3A - 6B) + 3(A+B) = 0 + 3(1)$
$=> -3B = 3$
$=> B = -1$

Then, substituting

$A - 1 = 1$
$=> A = 2$

So the decomposition is

$x/((x-6)(x-3)) = 2/(x-6) - 1/(x-3)$

Now we can do the integral relatively easily.

$intx/((x-6)(x-3))dx = int(2/(x-6) - 1/(x-3))dx$

$int(2/(x-6) - 1/(x-3))dx = 2int(1/(x-6))dx - int(1/(x-3))dx$

A couple of simple $u$ substitutions, together with $int(1/x)dx = ln|x|+C$
gives us

$2int(1/(x-6))dx = 2ln|x-6| + C$
and
$-int(1/(x-3))dx = -ln|x-3|+C$

So putting it all together, we get

$intx/((x-6)(x-3))dx = 2ln|x-6| - ln|x-3| + C$

How do you integrate int x/((x-6)(x-3) dx∫x(x−6)(x−3)dxint x/((x-6)(x-3) dx using partial fractions?