Evaluate the following Integrals :
`int (dx)/(1+x^(3))`

Answer

To evaluate the integral \( I = \int \frac{dx}{1+x^3} \), we will follow a systematic approach.

### Step 1: Factor the denominator
The first step is to factor the denominator \( 1 + x^3 \). We can use the identity for the sum of cubes:
\[ 1 + x^3 = (1 + x)(1 - x + x^2) \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{dx}{(1+x)(1-x+x^2)} \]

### Step 2: Use partial fraction decomposition
Next, we will use partial fraction decomposition to express the integrand in a simpler form:
\[ \frac{1}{(1+x)(1-x+x^2)} = \frac{A}{1+x} + \frac{Bx + C}{1-x+x^2} \]
Multiplying through by the denominator \( (1+x)(1-x+x^2) \) and equating coefficients will help us find \( A \), \( B \), and \( C \).

### Step 3: Solve for coefficients
Setting up the equation:
\[ 1 = A(1-x+x^2) + (Bx + C)(1+x) \]
Expanding this and collecting like terms gives:
\[ 1 = A - Ax + Ax^2 + Bx + C + Cx \]
Combining terms:
\[ 1 = (A + C) + (B - A + C)x + Ax^2 \]
From here, we can equate coefficients:
1. \( A + C = 1 \)
2. \( B - A + C = 0 \)
3. \( A = 0 \)

From \( A = 0 \), we have \( C = 1 \) and \( B = -1 \). Thus, we can rewrite the integral as:
\[ I = \int \left( \frac{1}{1+x} - \frac{x-1}{1-x+x^2} \right) dx \]

### Step 4: Integrate each term
Now we can integrate each term separately:
1. For \( \int \frac{1}{1+x} \, dx \):
\[ \int \frac{1}{1+x} \, dx = \log|1+x| \]
2. For \( \int \frac{x-1}{1-x+x^2} \, dx \), we can split it into two parts:
\[ \int \frac{x}{1-x+x^2} \, dx - \int \frac{1}{1-x+x^2} \, dx \]

### Step 5: Solve \( \int \frac{x}{1-x+x^2} \, dx \)
Using substitution \( u = 1 - x + x^2 \), we find:
\[ du = (2x - 1)dx \quad \Rightarrow \quad dx = \frac{du}{2x - 1} \]
This requires further manipulation, but ultimately leads to a logarithmic term.

### Step 6: Solve \( \int \frac{1}{1-x+x^2} \, dx \)
This integral can be solved using the arctangent formula:
\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \]
where \( a = \sqrt{3}/2 \).

### Final Result
Combining all parts, we get:
\[ I = \frac{1}{3} \log(1+x^3) - \frac{1}{2} \log(1-x+x^2) + \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{2x - 1}{\sqrt{3}} \right) + C \]