`int(sqrt(1+(logx)^2))/(x)dx=`

Options

A. `(logx)/(2)sqrt(1+(logx)^2)+(1)/(2)log|logx+sqrt(1+(logx)^2)|+c`
B. `(logx)/(2)sqrt(1+(logx)^2)-(1)/(2)log|logx+sqrt(1+(logx)^2)|+c`
C. `(logx)/(2)sqrt(1+(logx)^2)-(1)/(4)log|logx+sqrt(1+(logx)^2)|+c`
D. `(logx)/(2)sqrt(1+(logx)^2)+(1)/(4)log|logx+sqrt(1+(logx)^2)|+c`

Answer

To solve the integral \( I = \int \frac{\sqrt{1 + (\log x)^2}}{x} \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = \log x \). Then, the differential \( dt = \frac{1}{x} \, dx \) or \( dx = x \, dt = e^t \, dt \).

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we have:
\[ I = \int \sqrt{1 + t^2} \, dt \]

### Step 3: Integrate
The integral \( \int \sqrt{1 + t^2} \, dt \) can be solved using the formula:
\[ \int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \log |x + \sqrt{x^2 + a^2}| + C \]
In our case, \( a = 1 \) and \( x = t \):
\[ I = \frac{t}{2} \sqrt{t^2 + 1} + \frac{1}{2} \log |t + \sqrt{t^2 + 1}| + C \]

### Step 4: Substitute Back
Now we substitute back \( t = \log x \):
\[ I = \frac{\log x}{2} \sqrt{(\log x)^2 + 1} + \frac{1}{2} \log |\log x + \sqrt{(\log x)^2 + 1}| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = \frac{\log x}{2} \sqrt{(\log x)^2 + 1} + \frac{1}{2} \log |\log x + \sqrt{(\log x)^2 + 1}| + C \]

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