We are tasked with solving the integral:

$\int \frac{x}{2x^2 - 3x + 1} \, dx$

Steps to Solve:

1. Simplify the denominator if possible: The denominator $2x^2 - 3x + 1$ is a quadratic expression, and we should check if it can be factored.

   Factoring $2x^2 - 3x + 1$: $2x^2 - 3x + 1 = (2x - 1)(x - 1)$

   Thus, the integral becomes: $\int \frac{x}{(2x - 1)(x - 1)} \, dx$

2. Use partial fraction decomposition: Express the fraction as a sum of simpler fractions: $\frac{x}{(2x - 1)(x - 1)} = \frac{A}{2x - 1} + \frac{B}{x - 1}$ Multiply through by the denominator to find $A$ and $B$: $x = A(x - 1) + B(2x - 1)$ Expanding: $x = A(x) - A + B(2x) - B$ $x = (A + 2B)x - A - B$ Comparing coefficients of $x$ and the constant term:

   • Coefficient of $x$: $A + 2B = 1$
   • Constant term: $-A - B = 0$

   Solve the system of equations:

   1. $A + 2B = 1$
   2. $-A - B = 0$ (or $A = -B$)

   Substituting $A = -B$ into $A + 2B = 1$: $-B + 2B = 1 \implies B = 1$ $A = -1$

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

3. Integrate each term: $\int \frac{x}{(2x - 1)(x - 1)} \, dx = \int \frac{-1}{2x - 1} \, dx + \int \frac{1}{x - 1} \, dx$

   • First term: $\int \frac{-1}{2x - 1} \, dx = -\frac{1}{2} \ln|2x - 1|$

   • Second term: $\int \frac{1}{x - 1} \, dx = \ln|x - 1|$

4. Combine results: $\int \frac{x}{2x^2 - 3x + 1} \, dx = -\frac{1}{2} \ln|2x - 1| + \ln|x - 1| + C$

Final Answer:

$\boxed{-\frac{1}{2} \ln|2x - 1| + \ln|x - 1| + C}$

Would you like me to explain a specific step in more detail?

Related Questions:

1. How is partial fraction decomposition used in integration problems?
2. What happens if the denominator is not factorable?
3. Can this integral be solved using substitution instead of partial fractions?
4. How do you handle definite integrals with logarithmic terms?
5. How does the constant of integration affect the solution?

Tip:

Always check whether the denominator can be factored before jumping into other techniques; factoring often simplifies the problem significantly.