Step 1: Understanding the Question:
We need to solve the indefinite integral of a rational function where the degree of the numerator is higher than the denominator. Substitution is a good approach here. Step 2: Key Formula or Approach:
Substitute \( t = x + 1 \). Then \( x = t - 1 \) and \( dx = dt \). Step 3: Detailed Explanation:
Substitute into the integral:
\[ I = \int \frac{(t-1)^3}{t^2} dt \]
Expand the numerator using \( (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \):
\[ I = \int \frac{t^3 - 3t^2 + 3t - 1}{t^2} dt \]
Divide each term by \( t^2 \):
\[ I = \int \left( t - 3 + \frac{3}{t} - \frac{1}{t^2} \right) dt \]
Integrate term by term:
\[ I = \frac{t^2}{2} - 3t + 3 \log |t| - \left( -\frac{1}{t} \right) + C_1 \]
\[ I = \frac{t^2}{2} - 3t + 3 \log |t| + \frac{1}{t} + C_1 \]
Now substitute back \( t = x + 1 \):
\[ I = \frac{(x+1)^2}{2} - 3(x+1) + 3 \log |x+1| + \frac{1}{x+1} + C_1 \]
Expand the square and terms:
\[ I = \frac{x^2 + 2x + 1}{2} - 3x - 3 + 3 \log |x+1| + \frac{1}{x+1} + C_1 \]
\[ I = \frac{x^2}{2} + x + \frac{1}{2} - 3x - 3 + 3 \log |x+1| + \frac{1}{x+1} + C_1 \]
Combine the constants and linear terms:
\[ I = \frac{x^2}{2} - 2x + 3 \log |x+1| + \frac{1}{x+1} + \left( C_1 - \frac{5}{2} \right) \]
Let \( C_1 - \frac{5}{2} = c \):
\[ I = \frac{x^2}{2} - 2x + 3 \log(x+1) + \frac{1}{x+1} + c \] Step 4: Final Answer:
The integral result is \( \frac{x^2}{2} - 2x + 3 \log(x+1) + \frac{1}{x+1} + c \).