Evaluate `int (tanx+tan^3x)/(1+tan^3x)dx`

Answer

To evaluate the integral

\[ \int \frac{\tan x + \tan^3 x}{1 + \tan^3 x} \, dx, \]

we will follow these steps:

### Step 1: Simplify the Integral
We can factor out \(\tan x\) from the numerator:

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

Using the identity \(1 + \tan^2 x = \sec^2 x\), we rewrite the integral as:

\[ \int \frac{\tan x \sec^2 x}{1 + \tan^3 x} \, dx. \]

### Step 2: Substitution
Let \(t = \tan x\). Then, the derivative \(dt = \sec^2 x \, dx\). Therefore, we can rewrite the integral:

\[ \int \frac{t}{1 + t^3} \, dt. \]

### Step 3: Break Down the Integral
We can separate the integral by adding and subtracting \(t^2\):

\[ \int \frac{t + t^2 - t^2}{1 + t^3} \, dt = \int \frac{t}{1 + t^3} \, dt + \int \frac{t^2}{1 + t^3} \, dt. \]

### Step 4: Simplifying Further
Now, we can express the first integral:

\[ \int \frac{t}{1 + t^3} \, dt = \int \frac{t}{(1 + t)(1 - t + t^2)} \, dt. \]

### Step 5: Partial Fraction Decomposition
We can use partial fraction decomposition on \(\frac{t}{1 + t^3}\):

\[ \frac{t}{1 + t^3} = \frac{A}{1 + t} + \frac{Bt + C}{1 - t + t^2}. \]

### Step 6: Solve for Coefficients
Multiply through by the denominator \(1 + t^3\) and solve for \(A\), \(B\), and \(C\).

### Step 7: Integrate Each Term
After finding the coefficients, integrate each term separately. For example,

\[ \int \frac{A}{1 + t} \, dt = A \ln |1 + t| + C_1, \]
and for the quadratic term, use the formula for integrating rational functions.

### Step 8: Combine Results
Combine the results of the integrals and substitute back \(t = \tan x\).

### Final Result
The final result will be in terms of \(x\) and will include logarithmic terms and constants.