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}#