Integrate the following function:

$\int \frac{x \ dx}{(x-1)(x-2)}$ equals (A) $\log \left| \frac{(x-1)^2}{x-2} ight| + ext{C}$ (B) $\log \left| \frac{(x-2)^2}{x-1} ight| + ext{C}$ (C) $\log \left| \left( \frac{x-1}{x-2} ight)^2 ight| + ext{C}$ (D) $\log |(x-1)(x-2)

Question

Integrate the following function:

$\int \frac{x \ dx}{(x-1)(x-2)}$ equals (A) $\log \left| \frac{(x-1)^2}{x-2} ight| + 	ext{C}$ (B) $\log \left| \frac{(x-2)^2}{x-1} ight| + 	ext{C}$ (C) $\log \left| \left( \frac{x-1}{x-2} ight)^2 ight| + 	ext{C}$ (D) $\log |(x-1)(x-2)| + 	ext{C}$

Accepted Answer

> [!TLDR]
> We decompose the integrand into simpler fractions using partial fractions, integrate term by term, and combine the logs to match one of the given options. The result is $\log \left| \frac{(x-2)^2}{x-1} ight| + C$, which is option **(B)**.

The integrand $\frac{x}{(x-1)(x-2)}$ is a rational function where the denominator is already factored into distinct linear factors. Partial fraction decomposition lets us break this into a sum of simpler fractions, each with a single linear denominator. The reason this works is that the original fraction has a numerator of degree 1 and a denominator of degree 2 — so it's a proper fraction, and we can express it as:

$$
\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}
$$

where $A$ and $B$ are constants to be found. Once we find them, integrating becomes straightforward because $\int \frac{dx}{x-a} = \log|x-a| + C$.

Let's find $A$ and $B$.

1. **Set up the equation**  
   Multiply both sides by $(x-1)(x-2)$:
   $$
   x = A(x-2) + B(x-1)
   $$

2. **Solve for $A$ and $B$**  
   Expand the right side:
   $$
   x = A x - 2A + B x - B = (A+B)x + (-2A - B)
   $$
   Compare coefficients of $x$ and the constant term:
   $$
   \begin{cases}
   A + B = 1 \
   -2A - B = 0
   \end{cases}
   $$
   From the second equation, $B = -2A$. Substitute into the first:
   $$
   A - 2A = 1 \quad \Rightarrow \quad -A = 1 \quad \Rightarrow \quad A = -1
   $$
   Then $B = -2(-1) = 2$.

So:
   $$
   \frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2}
   $$

> [!TIP]
   > A faster way: for $A$, cover up $(x-1)$ in the denominator and evaluate the rest at $x=1$: $A = \frac{1}{1-2} = -1$. For $B$, cover up $(x-2)$ and evaluate at $x=2$: $B = \frac{2}{2-1} = 2$. This "cover-up" method works only for distinct linear factors.

3. **Integrate term by term**  
   $$
   \int \frac{x \, dx}{(x-1)(x-2)} = \int \left( \frac{-1}{x-1} + \frac{2}{x-2} ight) dx
   $$
   $$
   = -\int \frac{dx}{x-1} + 2 \int \frac{dx}{x-2}
   $$
   $$
   = -\lo

Q.Integrate the following function: $\int \frac{x \ dx}{(x-1)(x-2)}$ equals (A) $\log \left| \frac{(x-1)^2}{x-2} \right| + \text{C}$ (B) $\log \left| \frac{(x-2)^2}{x-1} \right| + \text{C}$ (C) $\log \left| \left( \frac{x-1}{x-2} \right)^2 \right| + \text{C}$ (D) $\log |(x-1)(x-2)| + \text{C}$

Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Method: Partial fractions, then fold logs into one to match options

Steps

Common Mistakes

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