`int 5/(1+x^4)dx`

Answer

To solve the integral \( I = \int \frac{5}{1 + x^4} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{5}{1 + x^4} \, dx \]

### Step 2: Factor the Denominator
Notice that \( 1 + x^4 \) can be factored as:
\[ 1 + x^4 = (1 + x^2)^2 - (x^2)^2 = (1 + x^2 - x)(1 + x^2 + x) \]
However, for simplicity, we will not factor it directly but will use a substitution later.

### Step 3: Use a Substitution
Let’s use the substitution \( x^2 = t \), hence \( dx = \frac{1}{2\sqrt{t}} \, dt \). The integral becomes:
\[ I = \int \frac{5}{1 + t^2} \cdot \frac{1}{2\sqrt{t}} \, dt \]

### Step 4: Simplify the Integral
Now we can rewrite the integral:
\[ I = \frac{5}{2} \int \frac{1}{1 + t^2} \cdot \frac{1}{\sqrt{t}} \, dt \]

### Step 5: Break Down the Integral
We can split the integral into two parts:
\[ I = \frac{5}{2} \left( \int \frac{1}{1 + t^2} \, dt - \int \frac{t}{1 + t^2} \, dt \right) \]

### Step 6: Solve Each Integral
1. The first integral:
\[ \int \frac{1}{1 + t^2} \, dt = \tan^{-1}(t) + C \]
2. The second integral can be solved using substitution \( u = 1 + t^2 \):
\[ \int \frac{t}{1 + t^2} \, dt = \frac{1}{2} \ln |1 + t^2| + C \]

### Step 7: Combine the Results
Putting it all together, we get:
\[ I = \frac{5}{2} \left( \tan^{-1}(t) - \frac{1}{2} \ln |1 + t^2| \right) + C \]

### Step 8: Substitute Back
Now substituting back \( t = x^2 \):
\[ I = \frac{5}{2} \left( \tan^{-1}(x^2) - \frac{1}{2} \ln |1 + x^4| \right) + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = \frac{5}{2} \tan^{-1}(x^2) - \frac{5}{4} \ln |1 + x^4| + C \]