If `int(x^2+4)/(x^4+16) dx=1/k tan^-1 ((x^2-4)/(kx))+c` then `k=` (i) `sqrt2` (ii)`4sqrt2` (iii)`2sqrt2` (iv) `2`

Options

A. 4
B. `2sqrt(2)`
C. 2
D. `sqrt(2)`

Explanation

To solve the integral

\[ \int \frac{x^2 + 4}{x^4 + 16} \, dx, \]

we will follow these steps:

### Step 1: Simplify the Integral
We start by dividing both the numerator and the denominator by \(x^2\):

\[ \int \frac{x^2 + 4}{x^4 + 16} \, dx = \int \frac{1 + \frac{4}{x^2}}{x^2 + \frac{16}{x^2}} \, dx. \]

### Step 2: Rewrite the Denominator
Next, we rewrite the denominator by adding and subtracting 8:

\[ x^4 + 16 = x^4 + 8 - 8 + 16 = (x^2)^2 + 8 + 8. \]

This allows us to express it in a form suitable for completing the square:

\[ x^4 + 16 = (x^2 - 4)^2 + 8. \]

### Step 3: Substitute
We will use the substitution \( t = x - \frac{4}{x} \). Differentiating gives:

\[ dt = \left(1 + \frac{4}{x^2}\right) dx. \]

Thus, we can express \(dx\) in terms of \(dt\):

\[ dx = \frac{dt}{1 + \frac{4}{x^2}}. \]

### Step 4: Change the Integral
Now we substitute \(t\) into the integral:

\[ \int \frac{dt}{t^2 + 8}. \]

### Step 5: Use the Integration Formula
We recognize that this integral can be solved using the formula:

\[ \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C. \]

Here, \(a = 2\sqrt{2}\):

\[ \int \frac{dt}{t^2 + (2\sqrt{2})^2} = \frac{1}{2\sqrt{2}} \tan^{-1} \left(\frac{t}{2\sqrt{2}}\right) + C. \]

### Step 6: Substitute Back
Substituting back for \(t\):

\[ = \frac{1}{2\sqrt{2}} \tan^{-1} \left(\frac{x - \frac{4}{x}}{2\sqrt{2}}\right) + C. \]

### Step 7: Final Form
This gives us:

\[ \int \frac{x^2 + 4}{x^4 + 16} \, dx = \frac{1}{2\sqrt{2}} \tan^{-1} \left(\frac{x^2 - 4}{2\sqrt{2}x}\right) + C. \]

### Step 8: Identify \(k\)
Comparing this with the given form

\[ \frac{1}{k} \tan^{-1} \left(\frac{x^2 - 4}{kx}\right) + C, \]

we find that \(k = 2\sqrt{2}\).

### Conclusion
Thus, the value of \(k\) is

\[ \boxed{2\sqrt{2}}. \]