How do you integrate $\int \frac{x^{3}}{x^{2} + 8 x + 16}$ $int (x^3)/(x^2 + 8x + 16)$ using partial fractions?

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How do you integrate $\int \frac{x^{3}}{x^{2} + 8 x + 16}$ $int (x^3)/(x^2 + 8x + 16)$ using partial fractions?

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Answer 1 · 2017-02-01T23:17:59

$\frac{1}{2} x^{2} - 8 x + 48 ln | x + 4 | + \frac{64}{x + 4} + c$ $1/2x^2-8x+48ln|x+4|+64/(x+4)+c$

Explanation:

The denominator is a perfect square, so you don't use partial fractions. Instead, you substitute $u = x + 4$ $u=x+4$ . The integral becomes:
$\int \frac{{(u - 4)}^{3}}{u^{2}} d u$ $int(u-4)^3/u^2 du$

$= \int \frac{u^{3} - 12 u^{2} + 48 u - 64}{u^{2}} d u$ $=int(u^3-12u^2+48u-64)/u^2du$

$= \int u - 12 + \frac{48}{u} - \frac{64}{u^{2}} d u$ $=intu-12+48/u-64/u^2du$

$= \frac{1}{2} u^{2} - 12 u + 48 ln | u | + \frac{64}{u} + k$ $=1/2u^2-12u+48ln|u|+64/u+k$

$= \frac{1}{2} {(x + 4)}^{2} - 12 (x + 4) + 48 ln | x + 4 | + \frac{64}{x + 4} + k$ $=1/2(x+4)^2-12(x+4)+48ln|x+4|+64/(x+4)+k$

$= \frac{1}{2} x^{2} - 8 x + 48 ln | x + 4 | + \frac{64}{x + 4} + c$ $=1/2x^2-8x+48ln|x+4|+64/(x+4)+c$

(where $c = 8 - 48 + k$ $c=8-48+k$ )

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How do you integrate $\int \frac{x^{3}}{x^{2} + 8 x + 16}$ $int (x^3)/(x^2 + 8x + 16)$ using partial fractions?

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Explanation:

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