Compute the following integrals.
(i) `int_(0)^(infty)f(x^(n)+x^(-n))lnx (dx)/(x)=0`
(ii) `int_(0)^(infty)f(x^(n)+x^(-n))lnx (dx)/(1+x^(2))=0`

Answer

To solve the given integrals, we will use a substitution method. Let's break down the solution step by step.

### Part (i)

We need to compute the integral:
\[ I = \int_{0}^{\infty} f(x^n + x^{-n}) \frac{\ln x}{x} \, dx \]

**Step 1: Substitution**
Let \( x = \frac{1}{t} \). Then, \( dx = -\frac{dt}{t^2} \). The limits change as follows:
- When \( x \to 0 \), \( t \to \infty \)
- When \( x \to \infty \), \( t \to 0 \)

Thus, the integral becomes:
\[ I = \int_{\infty}^{0} f\left(\left(\frac{1}{t}\right)^n + \left(\frac{1}{t}\right)^{-n}\right) \frac{\ln\left(\frac{1}{t}\right)}{\frac{1}{t}} \left(-\frac{dt}{t^2}\right) \]

**Step 2: Simplifying the Integral**
Rearranging the integral:
\[ I = \int_{0}^{\infty} f\left(\frac{1}{t^n} + t^n\right) \frac{-\ln t}{t} dt \]
Using the property \( \ln\left(\frac{1}{t}\right) = -\ln t \), we can write:
\[ I = \int_{0}^{\infty} f\left(t^{-n} + t^n\right) \frac{\ln t}{t} dt \]

**Step 3: Combining Integrals**
Now we have:
\[ I = -\int_{0}^{\infty} f\left(t^{-n} + t^n\right) \frac{\ln t}{t} dt \]
Let’s denote this new integral as \( I \) again. Thus, we have:
\[ I = -I \]

**Step 4: Solving for \( I \)**
Adding \( I \) to both sides:
\[ 2I = 0 \implies I = 0 \]

### Conclusion for Part (i)
The value of the integral is:
\[ \int_{0}^{\infty} f(x^n + x^{-n}) \frac{\ln x}{x} \, dx = 0 \]

---

### Part (ii)

Now we compute the second integral:
\[ J = \int_{0}^{\infty} f(x^n + x^{-n}) \frac{\ln x}{1 + x^2} \, dx \]

**Step 1: Substitution**
Again, let \( x = \frac{1}{t} \). Then, \( dx = -\frac{dt}{t^2} \) and the limits change as before:
\[ J = \int_{\infty}^{0} f\left(\left(\frac{1}{t}\right)^n + \left(\frac{1}{t}\right)^{-n}\right) \frac{\ln\left(\frac{1}{t}\right)}{1 + \left(\frac{1}{t}\right)^2} \left(-\frac{dt}{t^2}\right) \]

**Step 2: Simplifying the Integral**
Rearranging gives:
\[ J = \int_{0}^{\infty} f\left(t^{-n} + t^n\right) \frac{-\ln t}{1 + \frac{1}{t^2}} \frac{dt}{t^2} \]
This simplifies to:
\[ J = \int_{0}^{\infty} f\left(t^{-n} + t^n\right) \frac{\ln t}{1 + t^2} dt \]

**Step 3: Combining Integrals**
Now we have:
\[ J = -\int_{0}^{\infty} f\left(t^{-n} + t^n\right) \frac{\ln t}{1 + t^2} dt \]
Let’s denote this new integral as \( J \) again. Thus, we have:
\[ J = -J \]

**Step 4: Solving for \( J \)**
Adding \( J \) to both sides:
\[ 2J = 0 \implies J = 0 \]

### Conclusion for Part (ii)
The value of the integral is:
\[ \int_{0}^{\infty} f(x^n + x^{-n}) \frac{\ln x}{1 + x^2} \, dx = 0 \]

---