Problem 14 - Essential Calculus

Question

Find the indefinite integral. $$ \int \frac{x(x-2)}{(x-1)^{3}} d x $$

Step-by-Step Solution

Answer

\(\int \frac{x(x-2)}{(x-1)^{3}} dx = -\frac{1}{x-1} -\frac{1}{(x-1)^2} + C\)

Step 1: Simplifying the Function

It is essential to simplify the function before integrating it. The function given is of the form \(\frac{P(x)}{Q(x)}\), where both \(P\) and \(Q\) are polynomials. \(Q\) has higher degree than \(P\), so we can use polynomial division to simplify the function. In particular, we are able to rewrite \(x(x-2)\) as \(x^2-2x\), then divide \(x^2-2x\) by \((x-1)^{3}\). This allows us to rewrite \(x^2-2x\) as \((x-1)+(1)\) or \((x-1)\cdot1+1\). Then we find the remainder is 2. So, \(\frac{x(x-2)}{(x-1)^{3}} = \frac{1}{(x-1)^{2}} + \frac{2}{(x-1)^{3}}\).

Step 2: Integrating the Simplified Function

We integrate the resulting expressions individually: first, \(\int \frac{1}{(x-1)^{2}} dx = -\frac{1}{x-1}\) and \(\int \frac{2}{(x-1)^{3}} dx = -\frac{1}{(x-1)^2}\). Thus, the integral in the original problem is equal to the sum of these two integrals.

Step 3: Final Answer and Absolute Constant

Because we're solving an indefinite integral, we add a constant of integration at the end. Thus, our final answer is \(\int \frac{x(x-2)}{(x-1)^{3}} dx = -\frac{1}{x-1} -\frac{1}{(x-1)^2} + C\).