Evaluate: `int(e^tan^((-1)x))/((1+x^2))[((sec^(-1)sqrt(1+x^2))^2+cos^(-1)((1-x^2)/(1+x^2))]dx(x >0)dot`

Options

A. `e^(tan^(-1)x)*tan^(-1)x+C`
B. `(e^(tan^(-1)x)*(tan^(-1)x)^(2))/(2)+C`
C. `e^(tan^(-1)x)*(sec^(-1)(sqrt(1+x^(2))))^(2)+C`
D. `e^(tan^(-1)x)*(cosec^(-1)(sqrt(1+x^(2))))^(2)+C`

Answer

To evaluate the integral

\[ I = \int \frac{e^{\tan^{-1} x}}{1+x^2} \left[ \left( \sec^{-1} \sqrt{1+x^2} \right)^2 + \cos^{-1} \left( \frac{1-x^2}{1+x^2} \right) \right] dx \]

we can simplify the expression inside the integral using known identities.

### Step 1: Simplify the expression
Using the identities:
- \(\sec^{-1} \sqrt{1+x^2} = \tan^{-1} x\)
- \(\cos^{-1} \left( \frac{1-x^2}{1+x^2} \right) = 2 \tan^{-1} x\)

we can rewrite the integral as:

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

### Step 2: Substitute \(t = \tan^{-1} x\)
Let \(t = \tan^{-1} x\). Then, we have:

\[ x = \tan t \quad \text{and} \quad dx = \sec^2 t \, dt = (1 + \tan^2 t) \, dt = (1 + x^2) \, dt \]

Thus, the integral becomes:

\[ I = \int e^t (t^2 + 2t) dt \]

### Step 3: Apply integration by parts (Eilat's Rule)
Using Eilat's rule, we will integrate \(e^t (t^2 + 2t)\):

1. Let \(u = t^2 + 2t\) and \(dv = e^t dt\).
2. Then, \(du = (2t + 2) dt\) and \(v = e^t\).

Using integration by parts:

\[ I = e^t (t^2 + 2t) - \int e^t (2t + 2) dt \]

### Step 4: Repeat integration by parts
Now, we need to evaluate the integral \(\int e^t (2t + 2) dt\). We apply integration by parts again:

1. Let \(u = 2t + 2\) and \(dv = e^t dt\).
2. Then, \(du = 2 dt\) and \(v = e^t\).

Thus,

\[ \int e^t (2t + 2) dt = e^t (2t + 2) - \int 2 e^t dt \]

### Step 5: Evaluate the remaining integral
The remaining integral \(\int 2 e^t dt\) is straightforward:

\[ \int 2 e^t dt = 2 e^t \]

### Step 6: Combine results
Putting everything together, we have:

\[ I = e^t (t^2 + 2t) - \left( e^t (2t + 2) - 2 e^t \right) \]

Simplifying this gives:

\[ I = e^t (t^2 + 2t) - e^t (2t + 2) + 2 e^t \]

This simplifies to:

\[ I = e^t (t^2 + 2t - 2t - 2 + 2) = e^t t^2 \]

### Step 7: Substitute back to original variable
Substituting back \(t = \tan^{-1} x\):

\[ I = e^{\tan^{-1} x} \left( \tan^{-1} x \right)^2 + C \]

### Final Result
Thus, the final answer is:

\[ \int \frac{e^{\tan^{-1} x}}{1+x^2} \left[ \left( \sec^{-1} \sqrt{1+x^2} \right)^2 + \cos^{-1} \left( \frac{1-x^2}{1+x^2} \right) \right] dx = e^{\tan^{-1} x} \left( \tan^{-1} x \right)^2 + C \]