What is the antiderivative of $x^3/(x^4-5)^3$ ?

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Answer 1 · 2016-07-08T21:48:40

$= -1/8 * 1/(x^4-5)^2 color{green}{+ C}$

you can do this by a sub but i think that pattern spotting keeps it interesting/ enables you to do it in your head quite simply.

to start with , notice this pattern

$d/dx (1/(x^4-5)^2) = - 2 * 1/(x^4-5)^3 * 4x^3$

$= - 8 * x^3/(x^4-5)^3$

it's a mixture of the power rule and the chain rule.

and, so:

$d/dx (color{red}{-1/8} * 1/(x^4-5)^2) = x^3/(x^4-5)^3$

and by the FTC we have

$int dx qquad x^3/(x^4-5)^3 = -1/8 * 1/(x^4-5)^2 color{green}{+ C}$

What is the antiderivative of x^3/(x^4-5)^3x^3/(x^4-5)^3?