The integral `I=int2^((2^(x)+x))dx=lambda. (2^(2^(x)))+C` (where, C is the constant of integration). Then the value of `sqrtlambda` is equal to

Options

A. `(1)/(ln4)`
B. `(1)/((ln2)^(2))`
C. `(1)/(ln2)`
D. `(1)/((ln 4)^(2))`

Answer

To solve the integral \( I = \int 2^{2^x + x} \, dx \) and find the value of \( \sqrt{\lambda} \) where \( I = \lambda \cdot 2^{2x} + C \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int 2^{2^x + x} \, dx \]
We can rewrite the integrand:
\[ 2^{2^x + x} = 2^{2^x} \cdot 2^x \]
Thus, we have:
\[ I = \int 2^{2^x} \cdot 2^x \, dx \]

### Step 2: Substitution
Let \( t = 2^x \). Then, the derivative \( dt = 2^x \ln(2) \, dx \) implies:
\[ dx = \frac{dt}{t \ln(2)} \]
Now substituting \( t \) into the integral:
\[ I = \int t^2 \cdot \frac{dt}{t \ln(2)} = \frac{1}{\ln(2)} \int t \, dt \]

### Step 3: Integrate
Now we can integrate:
\[ \int t \, dt = \frac{t^2}{2} + C \]
Thus, substituting back:
\[ I = \frac{1}{\ln(2)} \cdot \left( \frac{t^2}{2} + C \right) = \frac{1}{2 \ln(2)} t^2 + C \]
Substituting \( t = 2^x \):
\[ I = \frac{1}{2 \ln(2)} (2^x)^2 + C = \frac{1}{2 \ln(2)} \cdot 2^{2x} + C \]

### Step 4: Compare with Given Form
We are given that:
\[ I = \lambda \cdot 2^{2x} + C \]
From our integration, we have:
\[ I = \frac{1}{2 \ln(2)} \cdot 2^{2x} + C \]
Comparing both expressions, we find:
\[ \lambda = \frac{1}{2 \ln(2)} \]

### Step 5: Find \( \sqrt{\lambda} \)
Now, we need to find \( \sqrt{\lambda} \):
\[ \sqrt{\lambda} = \sqrt{\frac{1}{2 \ln(2)}} \]
This can be simplified as:
\[ \sqrt{\lambda} = \frac{1}{\sqrt{2 \ln(2)}} \]

### Final Answer
Thus, the value of \( \sqrt{\lambda} \) is:
\[ \sqrt{\lambda} = \frac{1}{\sqrt{2 \ln(2)}} \]