To solve the integral [tex]\(\int \frac{dx}{(x-1)^5}\)[/tex], we will follow a systematic approach:
1. Substitution:
We start by making a substitution to simplify the integrand. Let [tex]\( u = x - 1 \)[/tex]. Then, the differential [tex]\( du = dx \)[/tex].
The integral now becomes:
[tex]\[
\int \frac{dx}{(x-1)^5} = \int \frac{du}{u^5}
\][/tex]
2. Integrate:
The above expression is a standard integral of the form [tex]\(\int u^n \, du\)[/tex], where [tex]\( n = -5 \)[/tex].
Recall the integral of [tex]\( u^n \)[/tex] is given by:
[tex]\[
\int u^n \, du = \frac{u^{n+1}}{n+1} + C
\][/tex]
provided [tex]\( n \neq -1 \)[/tex].
So, for our integral:
[tex]\[
\int \frac{1}{u^5} \, du = \int u^{-5} \, du
\][/tex]
Using the power rule:
[tex]\[
= \int u^{-5} \, du = \frac{u^{-5+1}}{-5+1} + C = \frac{u^{-4}}{-4} + C = -\frac{1}{4u^4} + C
\][/tex]
3. Substitute back [tex]\( u = x - 1 \)[/tex]:
We now revert our substitution to express the integral in terms of [tex]\( x \)[/tex]:
[tex]\[
-\frac{1}{4u^4} + C = -\frac{1}{4(x-1)^4} + C
\][/tex]
So the integral is:
[tex]\[
\int \frac{dx}{(x-1)^5} = -\frac{1}{4(x-1)^4} + C
\][/tex]
When expanding this answer to make it consistent with the format of the previously determined result, our steps will result in a different form for the representation, but both results inform the solution to the integration in distinct forms.
