Int (x/a+x^2)^3 dx = ?

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Int (x/a+x^2)^3 dx = ?

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Answer 1 · 2018-05-11T15:28:53

$(\frac{1}{7}) x^{7} + (\frac{1}{2 a}) x^{6} + (\frac{3}{5 a^{2}}) x^{5} + (\frac{1}{4 a^{3}}) x^{4} + C$ $(1/7)x^7 + (1/(2a))x^6 + (3/(5a^2))x^5 + (1/(4a^3))x^4 + C$

Explanation:

This integral can be solved by expanding it.

Note that the following is true:

${(\frac{x}{a} + x^{2})}^{3} = (\frac{x}{a} + x^{2}) (\frac{x}{a} + x^{2}) (\frac{x}{a} + x^{2})$ $(x/a + x^2)^3 = (x/a+x^2)(x/a + x^2)(x/a + x^2)$
$= (\frac{x^{2}}{a^{2}} + 2 \frac{x^{3}}{a} + x^{4}) (\frac{x}{a} + x^{2})$ $= (x^2/(a^2) + 2 x^3/a + x^4)(x/a + x^2)$
$= \frac{x^{3}}{a^{3}} + 2 \frac{x^{4}}{a^{2}} + \frac{x^{5}}{a} + \frac{x^{4}}{a^{2}} + 2 \frac{x^{5}}{a} + x^{6}$ $= x^3/(a^3) + 2x^4/(a^2) + x^5/a + x^4/(a^2) + 2x^5/a + x^6$
$= x^{6} + 3 \frac{x^{5}}{a} + 3 \frac{x^{4}}{a^{2}} + \frac{x^{3}}{a^{3}}$ $= x^6 + 3x^5/a + 3x^4 / (a^2) + x^3/(a^3)$

Making the above substitution, our integral is now:

$\int (x^{6} + (\frac{3}{a}) x^{5} + (\frac{3}{a^{2}}) x^{4} + (\frac{1}{a^{3}}) x^{3}) d x$ $int ( x^6 + (3/a)x^5 + (3/(a^2))x^4 + (1/(a^3))x^3)dx$
$= (\frac{1}{7}) x^{7} + (\frac{1}{2 a}) x^{6} + (\frac{3}{5 a^{2}}) x^{5} + (\frac{1}{4 a^{3}}) x^{4} + C$ $= (1/7)x^7 + (1/(2a))x^6 + (3/(5a^2))x^5 + (1/(4a^3))x^4 + C$

If it pleases you, you can find common denominators to make this a single term, but this form looks fine to me.

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Int (x/a+x^2)^3 dx = ?

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Explanation:

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