How do you integrate $\int \frac{6 (5 - x)}{(x - 7) (4 - x)}$ $int (6(5 - x))/( (x - 7)(4 - x))$ using partial fractions?

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How do you integrate $\int \frac{6 (5 - x)}{(x - 7) (4 - x)}$ $int (6(5 - x))/( (x - 7)(4 - x))$ using partial fractions?

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Answer 1 · 2017-04-13T18:41:53

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

Explanation:

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

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

By matching the numerators,

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

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

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

So, we have

$\int \frac{6 (5 - x)}{(x - 7) (4 - x)} d x = \int (\frac{4}{x - 7} + \frac{2}{x - 4}) d x$ $int(6(5-x))/((x-7)(4-x))dx =int(4/(x-7)+2/(x-4) )dx$

By Log Rule,

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

I hope that this was clear.

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How do you integrate $\int \frac{6 (5 - x)}{(x - 7) (4 - x)}$ $int (6(5 - x))/( (x - 7)(4 - x))$ using partial fractions?

1 Answer

Explanation:

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