How do you find $\int \frac{x - 4}{x^{2} - 4 x} d x$ $int (x - 4 ) / (x^2 -4x) dx$ using partial fractions?

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How do you find $\int \frac{x - 4}{x^{2} - 4 x} d x$ $int (x - 4 ) / (x^2 -4x) dx$ using partial fractions?

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Answer 1 · 2015-11-04T14:56:16

$\int \frac{x - 4}{x^{2} - 4 x} d x = ln | x |$ $int (x-4)/(x^2-4x)dx = ln |x|$

Explanation:

$\int \frac{x - 4}{x^{2} - 4 x} d x = \int \frac{x - 4}{x (x - 4)} d x = \int \frac{1}{x} d x = ln | x |$ $int (x-4)/(x^2-4x)dx = int (x-4)/(x(x-4))dx = int 1/x dx = ln |x|$

Answer 2 · 2015-11-04T15:02:04

Reduce the fraction.

Explanation:

When you factor the denominator to start the partial fraction decomposition, note that the ratio can be reduced.

$\int \frac{x - 4}{x^{2} - 4 x} d x = \int \frac{x - 4}{x (x - 4)} d x$ $int (x-4)/(x^2-4x) dx = int (x-4)/(x(x-4)) dx$

$= \int \frac{1}{x} d x = ln | x | + C$ $= int 1/x dx = lnabsx +C$

If you didn't notice it could be reduced, find $A, and B$ $A, "and " B$ so that:

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

So $A x - 4 A + B x = x - 4$ $Ax-4A+Bx = x-4$

And $A + B = 1$ $A+B=1$

and $- 4 A = - 4$ $-4A=-4$ , $" "$ so $A = 1$ $A=1$ and $B = 0$ $B=0$

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How do you find $\int \frac{x - 4}{x^{2} - 4 x} d x$ $int (x - 4 ) / (x^2 -4x) dx$ using partial fractions?

2 Answers

Explanation:

Explanation:

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How do you find $\int \frac{x - 4}{x^{2} - 4 x} d x$ $int (x - 4 ) / (x^2 -4x) dx$ using partial fractions?