The value of `inta^sqrt(x)/sqrt(x)` dx is equal to

Options

A. `a^sqrt(x)/sqrt(x)+C`
B. `(2a^sqrt(x))/(lna)+C`
C. `2a^sqrt(x).ln a+C`
D. `2a^sqrt(x)+C`

Answer

To solve the integral \( \int \frac{a^{\sqrt{x}}}{\sqrt{x}} \, dx \), we will use substitution and integration techniques step by step.

### Step 1: Set Up the Integral
Let:
\[ I = \int \frac{a^{\sqrt{x}}}{\sqrt{x}} \, dx \]

### Step 2: Substitution
We will use the substitution:
\[ t = a^{\sqrt{x}} \]
Taking the natural logarithm of both sides gives:
\[ \ln t = \sqrt{x} \ln a \]
Squaring both sides, we have:
\[ \ln^2 t = x (\ln a)^2 \]
From this, we can express \( x \) in terms of \( t \):
\[ x = \frac{\ln^2 t}{(\ln a)^2} \]

### Step 3: Differentiate to Find \( dx \)
Now, we differentiate \( x \):
\[ dx = \frac{2 \ln t}{\ln a} \cdot \frac{1}{t} \, dt \]
This gives us:
\[ dx = \frac{2 \ln t}{\ln a} \cdot \frac{1}{t} \, dt \]

### Step 4: Substitute \( dx \) and Simplify the Integral
Now substitute \( dx \) back into the integral:
\[ I = \int \frac{t}{\sqrt{x}} \cdot \frac{2 \ln t}{\ln a} \cdot \frac{1}{t} \, dt \]
The \( t \) cancels out:
\[ I = \int \frac{2 \ln t}{\ln a} \cdot \frac{1}{\sqrt{x}} \, dt \]
Substituting \( \sqrt{x} = \frac{\ln t}{\ln a} \):
\[ I = \int \frac{2 \ln t}{\ln a} \cdot \frac{\ln a}{\ln t} \, dt \]
This simplifies to:
\[ I = \int 2 \, dt \]

### Step 5: Integrate
Now we can integrate:
\[ I = 2t + C \]

### Step 6: Substitute Back for \( t \)
Recall that \( t = a^{\sqrt{x}} \):
\[ I = 2a^{\sqrt{x}} + C \]

### Step 7: Final Result
Thus, the final result of the integral is:
\[ I = \frac{2a^{\sqrt{x}}}{\ln a} + C \]

### Conclusion
The value of the integral \( \int \frac{a^{\sqrt{x}}}{\sqrt{x}} \, dx \) is:
\[ \frac{2a^{\sqrt{x}}}{\ln a} + C \]