Evaluate `int (e^(m tan^(-1)x))/(1+x^(2))dx`

Answer

To evaluate the integral

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

we will use substitution. Let's go through the steps:

### Step 1: Substitution
Let \( t = \tan^{-1} x \). Then, we differentiate to find \( dx \):
\[ \frac{dt}{dx} = \frac{1}{1 + x^2} \implies dx = (1 + x^2) \, dt. \]

### Step 2: Rewrite the Integral
Now, substitute \( t \) into the integral:
\[ \int \frac{e^{m \tan^{-1} x}}{1 + x^2} \, dx = \int e^{mt} \cdot \frac{1 + x^2}{1 + x^2} \, dt = \int e^{mt} \, dt. \]

### Step 3: Integrate
The integral of \( e^{mt} \) is:
\[ \int e^{mt} \, dt = \frac{e^{mt}}{m} + C, \]
where \( C \) is the constant of integration.

### Step 4: Substitute Back
Now, we need to substitute back \( t = \tan^{-1} x \):
\[ \frac{e^{m \tan^{-1} x}}{m} + C. \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \frac{e^{m \tan^{-1} x}}{1 + x^2} \, dx = \frac{e^{m \tan^{-1} x}}{m} + C. \]

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