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Evaluate the integral.$\int \frac{d x}{(x-4)^{2}(x-1)}$

$\int \frac{d x}{(x-4)^{2}(x-1)}=-\frac{1}{9}\left[\ln |x-4|+\frac{3}{(x-4)}-\ln |x-1|\right]+C$

Calculus 1 / AB

Calculus 2 / BC

Chapter 7

TECHNIQUES OF INTEGRATION

Section 5

The Method of Partial Fractions

Integration

Integration Techniques

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So we're gonna use decomposition of fractions to solve this. So we're gonna first start by decomposing our fraction. When we do this, we'll get a over X minus four plus B over X minus four squared plus C over X minus one. Then we multiply by the denominator, which is X minus four squared times X minus one, and we end up getting one equals a Times X minus four times X minus one plus B times X minus one plus C times X minus four squared then at X equals one we get that one equals by this should be 00 So we're just left with C times one minus four squared, which gives us C equals 1/9 and then X equals four. We get a that one equals it's going to get a 00 So we're just left with the times for minus one, which is going to be end up giving us the B equals one third. And then what? X equals zero. We end up getting that one equals for a minus are one third. That's RB value plus 16/9. Because that's not 16 times are see value. We're gonna end up getting a equals a negative 1/9. So finally, we can plug this back in two are integral, which is going to give us negative. Um, we'll take it out of the interest. Will have negative 1/9 of the integral one over X minus four class. Um, one third times the integral of X minus four to the negative, too DX dx, and then plus 1/9 of the integral from one over X minus one DX. So when we do all this, we end up getting as a result negative 1/9 times the natural log of absolute value of explains for plus three over X minus four time or minus the natural log of X minus one plus c Final answer.

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