How do you integrate $\int \frac{x^{3} - x^{2}}{{(x + 3)}^{4}}$ $int (x^3-x^2)/ (x+3)^4$ using partial fractions?

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How do you integrate $\int \frac{x^{3} - x^{2}}{{(x + 3)}^{4}}$ $int (x^3-x^2)/ (x+3)^4$ using partial fractions?

1 Answer

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Answer 1 · 2017-01-18T12:24:04

$ln | x + 3 | + \frac{10}{x + 3} - \frac{33}{2 {(x + 3)}^{2}} + \frac{12}{{(x + 3)}^{3}} + c$ $ln|x+3| +10/(x+3)-33/(2(x+3)^2)+12/(x+3)^3+c$

Explanation:

Let $u = x + 3$ $u=x+3$ , $\frac{d u}{d x} = 1$ $(du)/(dx)=1$ .
$\int \frac{{(u - 3)}^{3} - {(u - 3)}^{2}}{u^{4}} d u$ $int((u-3)^3-(u-3)^2)/u^4 du$
$= \int \frac{u^{3} - 10 u^{2} + 33 u - 36}{u^{4}} d u$ $=int(u^3-10u^2+33u-36)/u^4du$
$= \int u^{- 1} - 10 u^{- 2} + 33 u^{- 3} - 36 u^{- 4} d x$ $=int u^-1-10u^-2+33u^-3-36u^-4dx$
$= ln | u | + \frac{10}{u} - \frac{33}{2 u^{2}} + \frac{12}{u^{3}} + c$ $=ln|u|+10/u-33/(2u^2)+12/u^3+c$

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How do you integrate $\int \frac{x^{3} - x^{2}}{{(x + 3)}^{4}}$ $int (x^3-x^2)/ (x+3)^4$ using partial fractions?

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