How do you integrate $[\frac{6}{{(x - 3)}^{4}} \cdot (x^{2} - 4 x + 4)]$ $[6/(x-3)^4*(x^2-4x+4)]$ using partial fractions?

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How do you integrate $[\frac{6}{{(x - 3)}^{4}} \cdot (x^{2} - 4 x + 4)]$ $[6/(x-3)^4*(x^2-4x+4)]$ using partial fractions?

2 Answers

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Answer 1 · 2016-12-22T18:46:07

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Answer 2 · 2016-12-22T20:26:09

$\frac{- 6 x^{2} + 30 x - 38}{{(x - 3)}^{3}} + C$ $(-6x^2+30x-38)/(x-3)^3+C$

Explanation:

You don't. Instead, substitute $x = u + 3$ $x=u+3$ , $d x = d u$ $dx=du$ and the integral becomes $6 \int \frac{1}{u^{2}} + \frac{3}{u^{3}} + \frac{1}{u^{4}} d u$ $6 int 1/u^2+3/u^3+1/u^4du$
$= 6 (- \frac{1}{u} - \frac{1}{u^{2}} - \frac{\frac{1}{3}}{u^{3}}) + C$ $=6(-1/u-1/u^2-(1/3)/u^3)+C$ leading, upon substituting $u = x - 3$ $u=x-3$ , to the answer. This substitution always works well if the denominator is simply a power of a linear expression.

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How do you integrate $[\frac{6}{{(x - 3)}^{4}} \cdot (x^{2} - 4 x + 4)]$ $[6/(x-3)^4*(x^2-4x+4)]$ using partial fractions?

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How do you integrate [6(x−3)4⋅(x2−4x+4)][6/(x-3)^4*(x^2-4x+4)] using partial fractions?

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How do you integrate $[\frac{6}{{(x - 3)}^{4}} \cdot (x^{2} - 4 x + 4)]$ $[6/(x-3)^4*(x^2-4x+4)]$ using partial fractions?