To integrate the expression $\int \frac{x}{(x+1)^2} \, dx$, we can use substitution and some simplification.

Solution:

1. Substitute $u = x + 1$, which implies that $du = dx$ and $x = u - 1$.

   This substitution changes the integral as follows: $\int \frac{x}{(x+1)^2} \, dx = \int \frac{u - 1}{u^2} \, du$

2. Simplify the integrand: $= \int \left( \frac{u}{u^2} - \frac{1}{u^2} \right) \, du = \int \left( \frac{1}{u} - \frac{1}{u^2} \right) \, du$

3. Separate the integrals: $= \int \frac{1}{u} \, du - \int \frac{1}{u^2} \, du$

4. Integrate each term:

   • For $\int \frac{1}{u} \, du$, the result is $\ln |u|$.
   • For $\int \frac{1}{u^2} \, du$, rewrite it as $u^{-2}$, and integrate to get $-\frac{1}{u}$.

   So, $\int \frac{1}{u} \, du - \int \frac{1}{u^2} \, du = \ln |u| + \frac{1}{u} + C$

5. Substitute back $u = x + 1$: $= \ln |x + 1| + \frac{1}{x + 1} + C$

Final Answer:

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

Would you like more details on any step, or have any questions?

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Related Questions:

1. How would you integrate $\int \frac{x^2}{(x+1)^3} \, dx$?
2. Can you use a different method to integrate $\int \frac{x}{(x+1)^2} \, dx$?
3. What happens if we have $\int \frac{1}{(x+1)^2} \, dx$?
4. How do you find the antiderivative of $\int \frac{x}{(x+1)^n} \, dx$ for an arbitrary $n$?
5. What is the integral of $\int \frac{x+2}{(x+1)^2} \, dx$?

Tip:

When integrating rational functions, substitution is often helpful, especially when the derivative of the denominator closely resembles the numerator.