If `I=int_(0)^(16)(x^((1)/(4)))/(1+sqrtx)dx=k+4tan^(-1)m`, then `3k-m` is equal to

Answer

To solve the integral \( I = \int_{0}^{16} \frac{x^{\frac{1}{4}}}{1 + \sqrt{x}} \, dx \) and express it in the form \( k + 4 \tan^{-1} m \), we will follow these steps:

### Step 1: Substitution
Let \( x = t^4 \). Then, we differentiate to find \( dx \):
\[ dx = 4t^3 \, dt \]
Next, we change the limits of integration. When \( x = 0 \), \( t = 0 \), and when \( x = 16 \), \( t = 2 \) (since \( 16 = 2^4 \)).

### Step 2: Rewrite the Integral
Substituting \( x = t^4 \) into the integral, we get:
\[ I = \int_{0}^{2} \frac{(t^4)^{\frac{1}{4}}}{1 + \sqrt{t^4}} \cdot 4t^3 \, dt = \int_{0}^{2} \frac{t}{1 + t^2} \cdot 4t^3 \, dt \]
This simplifies to:
\[ I = 4 \int_{0}^{2} \frac{t^4}{1 + t^2} \, dt \]

### Step 3: Simplifying the Integral
We can separate the integrand:
\[ \frac{t^4}{1 + t^2} = \frac{t^4 - 1 + 1}{1 + t^2} = \frac{(t^2 - 1)(t^2 + 1)}{1 + t^2} + \frac{1}{1 + t^2} \]
Thus, we can rewrite the integral:
\[ I = 4 \left( \int_{0}^{2} (t^2 - 1) \, dt + \int_{0}^{2} \frac{1}{1 + t^2} \, dt \right) \]

### Step 4: Evaluate the Integrals
1. **First Integral**:
\[ \int_{0}^{2} (t^2 - 1) \, dt = \left[ \frac{t^3}{3} - t \right]_{0}^{2} = \left( \frac{8}{3} - 2 \right) - (0 - 0) = \frac{8}{3} - \frac{6}{3} = \frac{2}{3} \]

2. **Second Integral**:
\[ \int_{0}^{2} \frac{1}{1 + t^2} \, dt = \tan^{-1}(t) \bigg|_{0}^{2} = \tan^{-1}(2) - \tan^{-1}(0) = \tan^{-1}(2) \]

### Step 5: Combine the Results
Now substituting back into the expression for \( I \):
\[ I = 4 \left( \frac{2}{3} + \tan^{-1}(2) \right) = \frac{8}{3} + 4 \tan^{-1}(2) \]

### Step 6: Identify \( k \) and \( m \)
From the expression \( I = k + 4 \tan^{-1} m \):
- We can identify \( k = \frac{8}{3} \) and \( m = 2 \).

### Step 7: Calculate \( 3k - m \)
Now we calculate \( 3k - m \):
\[ 3k - m = 3 \cdot \frac{8}{3} - 2 = 8 - 2 = 6 \]

Thus, the final answer is:
\[ \boxed{6} \]