If ` lim_(a rarr infty) (1)/(a)int_(0)^(infty)(x^(2)+ax+1)/(1+x^(4)). "tan"^(-1)((1)/(x))dx ` is
equal to `(pi^(2))/(K),` where K in N, then K equals to

Options

A. 4
B. 8
C. 16
D. 32

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{a \to \infty} \frac{1}{a} \int_{0}^{\infty} \frac{x^2 + ax + 1}{1 + x^4} \tan^{-1}\left(\frac{1}{x}\right) dx \]

and find the value of \( K \) such that this limit equals \( \frac{\pi^2}{K} \).

### Step 1: Set up the integral

Let:

\[ I = \int_{0}^{\infty} \frac{x^2 + ax + 1}{1 + x^4} \tan^{-1}\left(\frac{1}{x}\right) dx \]

We need to evaluate:

\[ \lim_{a \to \infty} \frac{I}{a} \]

### Step 2: Use substitution

We perform the substitution \( x = \frac{1}{t} \), which gives \( dx = -\frac{1}{t^2} dt \). The limits change as follows: when \( x \to 0 \), \( t \to \infty \) and when \( x \to \infty \), \( t \to 0 \).

Thus, we have:

\[ I = \int_{\infty}^{0} \frac{\left(\frac{1}{t^2} + \frac{a}{t} + 1\right)}{1 + \frac{1}{t^4}} \tan^{-1}(t) \left(-\frac{1}{t^2}\right) dt \]

Rearranging gives:

\[ I = \int_{0}^{\infty} \frac{1 + at + t^2}{t^2(1 + \frac{1}{t^4})} \tan^{-1}(t) \frac{1}{t^2} dt \]

This simplifies to:

\[ I = \int_{0}^{\infty} \frac{1 + at + t^2}{1 + t^4} \tan^{-1}(t) dt \]

### Step 3: Combine integrals

Now we have two expressions for \( I \):

1. \( I = \int_{0}^{\infty} \frac{x^2 + ax + 1}{1 + x^4} \tan^{-1}\left(\frac{1}{x}\right) dx \)
2. \( I = \int_{0}^{\infty} \frac{1 + at + t^2}{1 + t^4} \tan^{-1}(t) dt \)

Adding these gives:

\[ 2I = \int_{0}^{\infty} \frac{(1 + ax + x^2)(\tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right))}{1 + x^4} dx \]

Using the identity \( \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \):

\[ 2I = \frac{\pi}{2} \int_{0}^{\infty} \frac{1 + ax + x^2}{1 + x^4} dx \]

### Step 4: Evaluate the integral

Now we need to evaluate:

\[ \int_{0}^{\infty} \frac{1 + ax + x^2}{1 + x^4} dx \]

This integral can be split into three parts:

\[ \int_{0}^{\infty} \frac{1}{1 + x^4} dx + a \int_{0}^{\infty} \frac{x}{1 + x^4} dx + \int_{0}^{\infty} \frac{x^2}{1 + x^4} dx \]

Using known results:

- \( \int_{0}^{\infty} \frac{1}{1 + x^4} dx = \frac{\pi}{2\sqrt{2}} \)
- \( \int_{0}^{\infty} \frac{x}{1 + x^4} dx = \frac{1}{2} \int_{0}^{\infty} \frac{1}{1 + u^2} du = \frac{\pi}{4} \)
- \( \int_{0}^{\infty} \frac{x^2}{1 + x^4} dx = \frac{\pi}{4\sqrt{2}} \)

### Step 5: Combine results

Thus,

\[ \int_{0}^{\infty} \frac{1 + ax + x^2}{1 + x^4} dx = \frac{\pi}{2\sqrt{2}} + a \cdot \frac{\pi}{4} + \frac{\pi}{4\sqrt{2}} \]

### Step 6: Substitute back into \( I \)

Substituting back into \( 2I \):

\[ 2I = \frac{\pi}{2} \left( \frac{\pi}{2\sqrt{2}} + a \cdot \frac{\pi}{4} + \frac{\pi}{4\sqrt{2}} \right) \]

### Step 7: Find limit as \( a \to \infty \)

Now we find:

\[ \lim_{a \to \infty} \frac{I}{a} = \lim_{a \to \infty} \frac{\frac{\pi^2}{4} a}{a} = \frac{\pi^2}{4} \]

### Step 8: Solve for \( K \)

We have:

\[ \lim_{a \to \infty} \frac{I}{a} = \frac{\pi^2}{K} \implies K = 4 \]

Thus, the final answer is:

\[ \boxed{4} \]