If `I_(1)=int_(x)^(1)(1)/(1+t^(2)) dt and I_(2)=int_(1)^(1//x)(1)/(1+t^(2)) dt "for" x gt0` then,

Options

A. `I_(1)=I_(2)`
B. `I_(1)gtI_(2)`
C. `I_(2)=I_(1)`
D. none of these

Explanation

To solve the problem, we need to find the relationship between the two integrals \( I_1 \) and \( I_2 \) defined as follows:

\[ I_1 = \int_{x}^{1} \frac{1}{1+t^2} \, dt \]
\[ I_2 = \int_{1}^{\frac{1}{x}} \frac{1}{1+t^2} \, dt \]

### Step 1: Calculate \( I_1 \)

To calculate \( I_1 \), we will use the formula for the integral of \( \frac{1}{1+t^2} \):

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

Now, we can evaluate \( I_1 \):

\[ I_1 = \int_{x}^{1} \frac{1}{1+t^2} \, dt = \tan^{-1}(t) \bigg|_{x}^{1} \]

Applying the limits:

\[ I_1 = \tan^{-1}(1) - \tan^{-1}(x) \]

Since \( \tan^{-1}(1) = \frac{\pi}{4} \):

\[ I_1 = \frac{\pi}{4} - \tan^{-1}(x) \]

### Step 2: Calculate \( I_2 \)

Next, we calculate \( I_2 \):

\[ I_2 = \int_{1}^{\frac{1}{x}} \frac{1}{1+t^2} \, dt = \tan^{-1}(t) \bigg|_{1}^{\frac{1}{x}} \]

Applying the limits:

\[ I_2 = \tan^{-1}\left(\frac{1}{x}\right) - \tan^{-1}(1) \]

Again, since \( \tan^{-1}(1) = \frac{\pi}{4} \):

\[ I_2 = \tan^{-1}\left(\frac{1}{x}\right) - \frac{\pi}{4} \]

### Step 3: Relate \( I_1 \) and \( I_2 \)

Now we have:

\[ I_1 = \frac{\pi}{4} - \tan^{-1}(x) \]
\[ I_2 = \tan^{-1}\left(\frac{1}{x}\right) - \frac{\pi}{4} \]

We can use the identity for the tangent inverse:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]

For \( a = x \) and \( b = \frac{1}{x} \):

\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}\left(\frac{x + \frac{1}{x}}{1 - 1}\right) = \frac{\pi}{2} \]

Thus, we can rewrite \( I_2 \):

\[ I_2 = \tan^{-1}\left(\frac{1}{x}\right) - \frac{\pi}{4} = \frac{\pi}{2} - \tan^{-1}(x) - \frac{\pi}{4} \]
\[ I_2 = \frac{\pi}{4} - \tan^{-1}(x) \]

### Conclusion

Now we can see that:

\[ I_1 = I_2 \]

Thus, the relationship between \( I_1 \) and \( I_2 \) is:

\[ I_1 = I_2 \]