Q. `int` `(ax^(n-1))/(bx^n + c)` `dx` =` ?`

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To take this integral, use substitution:

`u = bx^n + c`

Then `du = bnx^(n-1)dx` and from here

`dx=(du)/(bnx^(n-1))`

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`

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