Question

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

Answer 1

log|(x + 2)/(x + 1)| - 1/(1 + x) + c

Answer 2

To solve the integral \(\int \frac{1}{(x + 2)(1 + x)^2} \, dx\), we will use partial fraction decomposition.
First, decompose the integrand into partial fractions:
\[ \frac{1}{(x + 2)(1 + x)^2} = \frac{A}{x + 2} + \frac{B}{1 + x} + \frac{C}{(1 + x)^2} \]
Multiplying both sides by the common denominator \((x + 2)(1 + x)^2\) gives:
\[ 1 = A(1 + x)^2 + B(x + 2)(1 + x) + C(x + 2) \]
Expanding and simplifying the right side:
\[ 1 = A(1 + 2x + x^2) + B(x^2 + 3x + 2) + C(x + 2) \]
\[ 1 = A(x^2 + 2x + 1) + B(x^2 + 3x + 2) + Cx + 2C \]
Combine like terms:
\[ 1 = (A + B)x^2 + (2A + 3B + C)x + (A + 2B + 2C) \]
Now, equate the coefficients from both sides:
1. For \(x^2\): \(A + B = 0\)
2. For \(x\): \(2A + 3B + C = 0\)
3. For constant term: \(A + 2B + 2C = 1\)
Solving these equations:
1. \(A + B = 0\) implies \(B = -A\)
2. Substituting \(B = -A\) into \(2A + 3B + C = 0\):
\[ 2A + 3(-A) + C = 0 \]
\[ 2A - 3A + C = 0 \]
\[ -A + C = 0 \]
\[ C = A \]
3. Substituting \(B = -A\) and \(C = A\) into \(A + 2B + 2C = 1\):
\[ A + 2(-A) + 2(A) = 1 \]
\[ A - 2A + 2A = 1 \]
\[ A = 1 \]
So:
\[ A = 1 \]
\[ B = -1 \]
\[ C = 1 \]
Thus, the partial fractions decomposition is:
\[ \frac{1}{(x + 2)(1 + x)^2} = \frac{1}{x + 2} - \frac{1}{1 + x} + \frac{1}{(1 + x)^2} \]
Now integrate term-by-term:
\[ \int \frac{1}{(x + 2)(1 + x)^2} \, dx = \int \frac{1}{x + 2} \, dx - \int \frac{1}{1 + x} \, dx + \int \frac{1}{(1 + x)^2} \, dx \]
The integrals are:
1. \(\int \frac{1}{x + 2} \, dx = \ln|x + 2| + C_1\)
2. \(\int \frac{1}{1 + x} \, dx = \ln|1 + x| + C_2\)
3. \(\int \frac{1}{(1 + x)^2} \, dx = -\frac{1}{1 + x} + C_3\)
Combining these results, we get:
\[ \int \frac{1}{(x + 2)(1 + x)^2} \, dx = \ln|x + 2| - \ln|1 + x| - \frac{1}{1 + x} + C \]
Simplify using logarithm properties:
\[ \int \frac{1}{(x + 2)(1 + x)^2} \, dx = \ln \left| \frac{x + 2}{1 + x} \right| - \frac{1}{1 + x} + C \]
Therefore, the final answer is:
\[ \boxed{\ln \left| \frac{x + 2}{1 + x} \right| - \frac{1}{1 + x} + C} \]