How do you use Integration by Substitution to find #intdx/(1-6x)^4dx#?

1 Answer
Jacob F.
Aug 6, 2014

#int (1 - 6x)^(-4) dx =# ?

We will let #u = 1 - 6x#. Thus, #du = -6 dx#.

#= int u^(-4) dx#

This looks difficult since there isn't a #-6# in there for us to form a #du# with. However, there is a rule of integration which states:

#int c*f(x) dx = c * int f(x) dx#

We can exploit this rule to rewrite our integral equivalently as:

#= -1/6 int -6 u^(-4) dx#

The statements are completely equivalent; note that if we pull the #-6# out of the integral, then it cancels with the #-1/6# and leaves us with #1# multiplied by our original integral.

Anyway, we now have a #-6 dx# to form a #du# with.

#= -1/6 int u^(-4) du#

#= -1/6 u^(-3) * (-1/3)#

#= 1/18 u^(-3)#

#= 1/(18u^3)#

Substituting back for #u# gives us:

#= 1/(18(1 - 6x)^3)#

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