`(i) int (1)/(x(x+1)^(2))dx`
`(ii) int (1)/((x+1)^(2)(x-1))dx`

Answer

Let's solve the given integrals step by step.

### Part (i):
**Integral to solve:**
\[ \int \frac{1}{x(x+1)^2} \, dx \]

**Step 1: Partial Fraction Decomposition**
We start by expressing the integrand in terms of partial fractions:
\[ \frac{1}{x(x+1)^2} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \]
Multiplying through by the denominator \(x(x+1)^2\) gives:
\[ 1 = A(x+1)^2 + Bx(x+1) + Cx \]

**Step 2: Expand and Collect Terms**
Expanding the right-hand side:
\[ 1 = A(x^2 + 2x + 1) + B(x^2 + x) + Cx \]
\[ = (A + B)x^2 + (2A + B + C)x + A \]

**Step 3: Set Up System of Equations**
Now, we equate coefficients from both sides:
1. \(A + B = 0\) (coefficient of \(x^2\))
2. \(2A + B + C = 0\) (coefficient of \(x\))
3. \(A = 1\) (constant term)

From the third equation, we find \(A = 1\). Substituting \(A\) into the first equation:
\[ 1 + B = 0 \implies B = -1 \]
Substituting \(A\) and \(B\) into the second equation:
\[ 2(1) - 1 + C = 0 \implies 2 - 1 + C = 0 \implies C = -1 \]

Thus, we have:
\[ A = 1, \quad B = -1, \quad C = -1 \]

**Step 4: Rewrite the Integral**
Substituting back into the partial fraction decomposition:
\[ \int \left( \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2} \right) \, dx \]

**Step 5: Integrate Each Term**
Now we integrate term by term:
\[ = \int \frac{1}{x} \, dx - \int \frac{1}{x+1} \, dx - \int \frac{1}{(x+1)^2} \, dx \]
This gives:
\[ = \ln |x| - \ln |x+1| + \frac{1}{x+1} + C \]

**Final Result for Part (i):**
\[ \ln \left| \frac{x}{x+1} \right| + \frac{1}{x+1} + C \]

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### Part (ii):
**Integral to solve:**
\[ \int \frac{1}{(x+1)^2(x-1)} \, dx \]

**Step 1: Partial Fraction Decomposition**
We express the integrand in terms of partial fractions:
\[ \frac{1}{(x+1)^2(x-1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \]
Multiplying through by the denominator \((x+1)^2(x-1)\) gives:
\[ 1 = A(x+1)^2 + B(x-1)(x+1) + C(x-1) \]

**Step 2: Expand and Collect Terms**
Expanding the right-hand side:
\[ 1 = A(x^2 + 2x + 1) + B(x^2 - 1) + C(x - 1) \]
\[ = (A + B)x^2 + (2A + C)x + (A - C) \]

**Step 3: Set Up System of Equations**
Equating coefficients:
1. \(A + B = 0\) (coefficient of \(x^2\))
2. \(2A + C = 0\) (coefficient of \(x\))
3. \(A - C = 1\) (constant term)

From the first equation, \(B = -A\). Substituting into the second:
\[ 2A + C = 0 \implies C = -2A \]
Substituting \(C\) into the third equation:
\[ A - (-2A) = 1 \implies 3A = 1 \implies A = \frac{1}{3} \]
Then, substituting back:
\[ B = -\frac{1}{3}, \quad C = -\frac{2}{3} \]

**Step 4: Rewrite the Integral**
Substituting back into the partial fraction decomposition:
\[ \int \left( \frac{1/3}{x-1} - \frac{1/3}{x+1} - \frac{2/3}{(x+1)^2} \right) \, dx \]

**Step 5: Integrate Each Term**
Now we integrate term by term:
\[ = \frac{1}{3} \int \frac{1}{x-1} \, dx - \frac{1}{3} \int \frac{1}{x+1} \, dx - \frac{2}{3} \int \frac{1}{(x+1)^2} \, dx \]
This gives:
\[ = \frac{1}{3} \ln |x-1| - \frac{1}{3} \ln |x+1| + \frac{2}{3(x+1)} + C \]

**Final Result for Part (ii):**
\[ \frac{1}{3} \ln \left| \frac{x-1}{x+1} \right| + \frac{2}{3(x+1)} + C \]

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