Prove that `int e^x (tan^-1 x+1/(1+x^2)) dx=e^x tan^-1x+c`

Answer

To prove that

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

we will use integration by parts and the properties of derivatives.

### Step 1: Identify the functions for integration by parts
Let:
- \( u = \tan^{-1} x \)
- \( dv = e^x dx \)

### Step 2: Differentiate and integrate
Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{1+x^2} dx \]
- Integrate \( dv \):
\[ v = e^x \]

### Step 3: Apply integration by parts formula
The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]

Substituting our values into this formula:
\[ \int e^x \tan^{-1} x \, dx = e^x \tan^{-1} x - \int e^x \cdot \frac{1}{1+x^2} \, dx \]

### Step 4: Combine the integrals
Now we can rewrite the original integral:
\[ \int e^x \left( \tan^{-1} x + \frac{1}{1+x^2} \right) dx = \int e^x \tan^{-1} x \, dx + \int e^x \cdot \frac{1}{1+x^2} \, dx \]

Substituting the result from the integration by parts:
\[ = \left( e^x \tan^{-1} x - \int e^x \cdot \frac{1}{1+x^2} \, dx \right) + \int e^x \cdot \frac{1}{1+x^2} \, dx \]

### Step 5: Simplify the expression
Notice that the two integrals \( \int e^x \cdot \frac{1}{1+x^2} \, dx \) cancel out:
\[ = e^x \tan^{-1} x + C \]

### Conclusion
Thus, we have shown that:
\[ \int e^x \left( \tan^{-1} x + \frac{1}{1+x^2} \right) dx = e^x \tan^{-1} x + C \]