Evaluate :
(i) `int_(-oo)^(oo) (dx)/(x^(2)+2x+2)` , (ii) `int_(sqrt(2))^(oo)(dx)/(xsqrt(x^(2)-1))`, (iii) `int_(0)^(4)(x^(2))/(1+x)dx`
(iv) `int_(0)^(pi//2) sqrt(costheta)sin^(3)theta`

Answer

Let's evaluate the integrals step by step.

### (i) Evaluate \( \int_{-\infty}^{\infty} \frac{dx}{x^2 + 2x + 2} \)

1. **Complete the square**:
\[ x^2 + 2x + 2 = (x + 1)^2 + 1 \]
So, we rewrite the integral:
\[ \int_{-\infty}^{\infty} \frac{dx}{(x + 1)^2 + 1} \]

2. **Use the formula for integration**:
The integral of \( \frac{1}{x^2 + a^2} \) is \( \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \).
Here, \( a = 1 \), so:
\[ \int \frac{dx}{(x + 1)^2 + 1} = \tan^{-1}(x + 1) \]

3. **Evaluate the limits**:
\[ \left[ \tan^{-1}(x + 1) \right]_{-\infty}^{\infty} = \tan^{-1}(\infty) - \tan^{-1}(-\infty) = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi \]

Thus, the result for (i) is:
\[ \int_{-\infty}^{\infty} \frac{dx}{x^2 + 2x + 2} = \pi \]

### (ii) Evaluate \( \int_{\sqrt{2}}^{\infty} \frac{dx}{x \sqrt{x^2 - 1}} \)

1. **Use the known integral**:
The integral \( \int \frac{dx}{x \sqrt{x^2 - 1}} = \cos^{-1}(x) + C \).

2. **Evaluate the limits**:
\[ \left[ \cos^{-1}(x) \right]_{\sqrt{2}}^{\infty} = \cos^{-1}(\infty) - \cos^{-1}(\sqrt{2}) \]
Here, \( \cos^{-1}(\infty) = 0 \) and \( \cos^{-1}(\sqrt{2}) = \frac{\pi}{4} \).

Thus, the result for (ii) is:
\[ \int_{\sqrt{2}}^{\infty} \frac{dx}{x \sqrt{x^2 - 1}} = -\frac{\pi}{4} \]

### (iii) Evaluate \( \int_{0}^{4} \frac{x^2}{1 + x} dx \)

1. **Split the integral**:
\[ \int_{0}^{4} \frac{x^2}{1 + x} dx = \int_{0}^{4} \left( x - 1 + \frac{1}{1 + x} \right) dx \]

2. **Integrate each term**:
\[ \int_{0}^{4} x \, dx - \int_{0}^{4} 1 \, dx + \int_{0}^{4} \frac{1}{1 + x} \, dx \]
- \( \int_{0}^{4} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{16}{2} = 8 \)
- \( \int_{0}^{4} 1 \, dx = [x]_{0}^{4} = 4 \)
- \( \int_{0}^{4} \frac{1}{1 + x} \, dx = \left[ \ln(1 + x) \right]_{0}^{4} = \ln(5) - \ln(1) = \ln(5) \)

3. **Combine the results**:
\[ 8 - 4 + \ln(5) = 4 + \ln(5) \]

Thus, the result for (iii) is:
\[ \int_{0}^{4} \frac{x^2}{1 + x} dx = 4 + \ln(5) \]

### (iv) Evaluate \( \int_{0}^{\frac{\pi}{2}} \sqrt{\cos \theta} \sin^3 \theta \, d\theta \)

1. **Rewrite the integral**:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{\cos \theta} \sin^3 \theta \, d\theta = \int_{0}^{\frac{\pi}{2}} \sqrt{\cos \theta} (1 - \cos^2 \theta) \sin \theta \, d\theta \]

2. **Split the integral**:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{\cos \theta} \sin \theta \, d\theta - \int_{0}^{\frac{\pi}{2}} \sqrt{\cos \theta} \cos^2 \theta \sin \theta \, d\theta \]

3. **Substitute \( u = \cos \theta \)**:
- For the first integral, \( du = -\sin \theta \, d\theta \):
\[ = -\int_{1}^{0} u^{1/2} \, du = \int_{0}^{1} u^{1/2} \, du = \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{1} = \frac{2}{3} \]

- For the second integral:
\[ = -\int_{1}^{0} u^{5/2} \, du = \int_{0}^{1} u^{5/2} \, du = \left[ \frac{u^{7/2}}{7/2} \right]_{0}^{1} = \frac{2}{7} \]

4. **Combine the results**:
\[ \frac{2}{3} - \frac{2}{7} = \frac{14 - 6}{21} = \frac{8}{21} \]

Thus, the result for (iv) is:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{\cos \theta} \sin^3 \theta \, d\theta = \frac{8}{21} \]

### Summary of Results:
1. \( \int_{-\infty}^{\infty} \frac{dx}{x^2 + 2x + 2} = \pi \)
2. \( \int_{\sqrt{2}}^{\infty} \frac{dx}{x \sqrt{x^2 - 1}} = -\frac{\pi}{4} \)
3. \( \int_{0}^{4} \frac{x^2}{1 + x} dx = 4 + \ln(5) \)
4. \( \int_{0}^{\frac{\pi}{2}} \sqrt{\cos \theta} \sin^3 \theta \, d\theta = \frac{8}{21} \)