To solve the integral:

$\int \frac{2x}{(x - 1)^2} \, dx$

we can proceed by using substitution.

1. Choose a substitution: Let $u = x - 1$. Then, $du = dx$ and $x = u + 1$.

2. Rewrite the integrand in terms of $u$: Substituting $x = u + 1$ into the integrand: $\frac{2x}{(x - 1)^2} = \frac{2(u + 1)}{u^2}.$

   So, the integral becomes: $\int \frac{2(u + 1)}{u^2} \, du.$

3. Simplify the expression: Split the fraction: $\int \frac{2(u + 1)}{u^2} \, du = \int \left( \frac{2u}{u^2} + \frac{2}{u^2} \right) \, du = \int \left( \frac{2}{u} + \frac{2}{u^2} \right) \, du.$

4. Integrate each term: Now integrate term-by-term: $\int \frac{2}{u} \, du + \int \frac{2}{u^2} \, du.$

   • The first term, $\int \frac{2}{u} \, du$, integrates to $2 \ln |u|$.
   • The second term, $\int \frac{2}{u^2} \, du$, integrates to $-\frac{2}{u}$.

   So, we get: $2 \ln |u| - \frac{2}{u}.$

5. Substitute back $u = x - 1$: Replacing $u$ with $x - 1$, the result is: $2 \ln |x - 1| - \frac{2}{x - 1} + C,$ where $C$ is the constant of integration.

Final Answer:

$\int \frac{2x}{(x - 1)^2} \, dx = 2 \ln |x - 1| - \frac{2}{x - 1} + C.$

Would you like a deeper explanation on any step, or have any questions?

Here are 5 related questions you might find helpful:

1. How would you approach this integral if the power in the denominator were different?
2. What other techniques could we use to solve integrals with similar fractions?
3. How does substitution simplify complex rational functions?
4. How would the solution change if the numerator were $x^2$ instead of $2x$?
5. How can you verify the answer by differentiating the result?

Tip: When integrating rational functions, consider substitution if you see an expression and its derivative (or close to it) in the integrand.