` (i) int(1)/((1+x^(2))tan ^(-1) x )dx " "(ii) int(e^(tan^(-1)x))/(1+x^(2))dx`

Answer

Let's solve the integrals step by step.

### Part (i): Evaluate \( I_1 = \int \frac{1}{(1+x^2) \tan^{-1} x} \, dx \)

1. **Substitution**: Let \( t = \tan^{-1} x \).
- Then, differentiate both sides:
\[ \frac{dt}{dx} = \frac{1}{1+x^2} \implies dx = (1+x^2) dt \]
- Substitute \( dx \) and \( \tan^{-1} x \) in the integral:
\[ I_1 = \int \frac{1}{(1+x^2) t} (1+x^2) dt = \int \frac{1}{t} dt \]

2. **Integrate**:
- The integral of \( \frac{1}{t} \) is:
\[ I_1 = \ln |t| + C \]

3. **Back-substitution**:
- Replace \( t \) with \( \tan^{-1} x \):
\[ I_1 = \ln |\tan^{-1} x| + C \]

### Final Result for Part (i):
\[ I_1 = \ln |\tan^{-1} x| + C \]

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### Part (ii): Evaluate \( I_2 = \int \frac{e^{\tan^{-1} x}}{1+x^2} \, dx \)

1. **Substitution**: Again, let \( t = \tan^{-1} x \).
- As before, we have:
\[ dx = (1+x^2) dt \]
- Substitute \( dx \) and \( \tan^{-1} x \) in the integral:
\[ I_2 = \int e^t \cdot dt \]

2. **Integrate**:
- The integral of \( e^t \) is:
\[ I_2 = e^t + C_1 \]

3. **Back-substitution**:
- Replace \( t \) with \( \tan^{-1} x \):
\[ I_2 = e^{\tan^{-1} x} + C_1 \]

### Final Result for Part (ii):
\[ I_2 = e^{\tan^{-1} x} + C_1 \]

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