Question

Evaluate the integral \[ \int \frac{x^3}{(x+1)^2} \, dx. \]

• A. \( \frac{x^2}{3} + 3 \log(x + 1) + c \)
• B. \( \frac{x^2}{2} + 3 \log(x + 1) + c \)
• C. \( \frac{x^2}{3} + \log(x + 1) + c \)
• D. \( \frac{x^2}{2} + \log(x + 1) + c \)

Hint:

When dealing with polynomials over powers of binomials, use substitution and expand the terms to simplify the integral.

Answer 1

The Correct Option is B

Step 1: Applying substitution.
We are given the integral:
\[ \int \frac{x^3}{(x+1)^2} \, dx. \]
First, we perform a substitution to simplify the expression. Let:
\[ u = x + 1 \quad \Rightarrow \quad du = dx, \quad x = u - 1. \]
Substituting these into the integral: \[ \int \frac{(u-1)^3}{u^2} \, du. \]

Step 2: Expanding the numerator.
Expanding \( (u-1)^3 \): \[ (u-1)^3 = u^3 - 3u^2 + 3u - 1. \]
Thus, the integral becomes:
\[ \int \frac{u^3 - 3u^2 + 3u - 1}{u^2} \, du = \int (u - 3 + \frac{3}{u} - \frac{1}{u^2}) \, du. \]

Step 3: Integrating term by term.
Now, integrate each term:
\[ \int u \, du = \frac{u^2}{2}, \quad \int -3 \, du = -3u, \quad \int \frac{3}{u} \, du = 3 \log |u|, \quad \int -\frac{1}{u^2} \, du = \frac{1}{u}. \]

Step 4: Substituting back.
Substituting \( u = x + 1 \) back into the result:
\[ \frac{(x+1)^2}{2} - 3(x + 1) + 3 \log |x + 1| + \frac{1}{x+1}. \]

Step 5: Final simplification.
Simplifying the expression gives:
\[ \frac{x^2}{2} + 3 \log(x + 1) + c. \]
Final Answer:
Thus, the correct answer is:
\[ \boxed{\frac{x^2}{2} + 3 \log(x + 1) + c}. \]