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

Answer

To solve the integral

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

we can use a substitution method. Let's go through the 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
Substituting \( t \) into the integral, we have:

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

Thus, the integral becomes:

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

### Step 3: Integrate
Now, we can integrate:

\[ \int e^{m t} \, dt = \frac{1}{m} e^{m t} + C, \]

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

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

\[ \frac{1}{m} e^{m \tan^{-1} x} + C. \]

### Final Answer
Thus, the final result of the integral is:

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

---