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

Question

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

Samantha Adkison · Accepted Answer

#int(6(5-x))/((x-7)(4-x)) dx=4ln|x-7|+2ln|x-4|+C#

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

#(6(x-5))/((x-7)(x-4))=A/(x-7)+B/(x-4)=(A(x-4)+B(x-7))/((x-7)(x-4))#

When the numerators coincide,

#A(x-4)+B(x-7)=6(x-5)#

To find #A#, set #x=7 Rightarrow 3A=12 Rightarrow A=4#

To find #B#, set #x=4 Rightarrow -3B=-6 Rightarrow B=2#

Thus, we have

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

According to Log Rule,

#=4ln|x-7|+2ln|x-4|+C#

I hope this made sense.

Paisley Johnson · Answer

undefined To integrate $ \int \frac{{6(5 - x)}}{{(x - 7)(4 - x)}} $ using partial fractions, you'll first decompose the fraction into simpler fractions using partial fraction decomposition.

1. Decompose $ \frac{{6(5 - x)}}{{(x - 7)(4 - x)}} $ into partial fractions.
2. Find the constants $ A $ and $ B $ such that $ \frac{{6(5 - x)}}{{(x - 7)(4 - x)}} = \frac{A}{{x - 7}} + \frac{B}{{4 - x}} $.
3. Clear the fractions and solve for $ A $ and $ B $.
4. Integrate each partial fraction term separately.

Following these steps:

$$
\begin{align*}
\frac{{6(5 - x)}}{{(x - 7)(4 - x)}} &= \frac{A}{{x - 7}} + \frac{B}{{4 - x}} \
6(5 - x) &= A(4 - x) + B(x - 7) \
&= (A - B)x + (4A - 7B)
\end{align*}
$$

Matching coefficients:
$$ 5(6) = 4A - 7B $$
$$ -6 = A - B $$

Solving these equations gives $ A = -1 $ and $ B = -5 $.

So, $ \frac{{6(5 - x)}}{{(x - 7)(4 - x)}} = \frac{{-1}}{{x - 7}} + \frac{{-5}}{{4 - x}} $.

Now, integrate each term separately:
$$ \int \frac{{-1}}{{x - 7}} \, dx = -\ln|x - 7| + C_1 $$
$$ \int \frac{{-5}}{{4 - x}} \, dx = -5\ln|4 - x| + C_2 $$

Where $ C_1 $ and $ C_2 $ are constants of integration.

So, the integral of $ \frac{{6(5 - x)}}{{(x - 7)(4 - x)}} $ is:
$$ -\ln|x - 7| - 5\ln|4 - x| + C $$
where $ C = C_1 + C_2 $.