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Q. `int` `(ax^(n-1))/(bx^n + c)` `dx` =` ?`
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Best answer as selected by question asker.
To take this integral, use substitution:
`u = bx^n + c`
Then `du = bnx^(n-1)dx` and from here
Plugging this in the integral, get
`int (ax^(n-1))/(bx^n+c) dx= int a/(bn)(du)/u = a/(bn) int (du)/u`
Antiderivative of `1/u` ls ln(u), so
`int (du)/u = lnu + constant`
and the original integral equals `a/(bn) ln(u)+constant`
Substitute the expression for u back:
`int (ax^(n-1))/(bx^n+c) dx= a/(bn) ln(bx^n + c)+ constant`
Posted by ishpiro on July 12, 2013 at 1:38 AM (Answer #1)
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