Evaluate the following integrals:
`int e^(tan^-1x)times[frac{1}{1 +x^2}]dx`

Answer

To evaluate the integral

\[ I = \int e^{\tan^{-1} x} \cdot \frac{1}{1 + x^2} \, dx, \]

we will follow these steps:

### Step 1: Substitution
Let \( t = \tan^{-1} x \). Then, we differentiate both sides with respect to \( x \):

\[ \frac{dt}{dx} = \frac{1}{1 + x^2} \implies dx = (1 + x^2) \, dt. \]

### Step 2: Rewrite the Integral
Now, substituting \( t \) into the integral, we have:

\[ I = \int e^t \cdot \frac{1}{1 + x^2} \cdot (1 + x^2) \, dt. \]

Notice that \( \frac{1}{1 + x^2} \cdot (1 + x^2) = 1 \). Therefore, the integral simplifies to:

\[ I = \int e^t \, dt. \]

### Step 3: Integrate
The integral of \( e^t \) is:

\[ \int e^t \, dt = e^t + C, \]

where \( C \) is the constant of integration.

### Step 4: Back Substitute
Now, we substitute back \( t = \tan^{-1} x \):

\[ I = e^{\tan^{-1} x} + C. \]

### Final Answer
Thus, the evaluated integral is:

\[ \int e^{\tan^{-1} x} \cdot \frac{1}{1 + x^2} \, dx = e^{\tan^{-1} x} + C. \]

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