The value of `int_(0)^(4) 3^(sqrt(2x+1) )dx` is

Options

A. `(6)/(log 3) (13- (4)/(log 3) )`
B. `(66)/(log 3)`
C. `(6)/(log 3) (13- (5)/(log 3) )`
D. none of these

Answer

To solve the integral \( I = \int_{0}^{4} 3^{\sqrt{2x+1}} \, dx \), we will follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{0}^{4} 3^{\sqrt{2x+1}} \, dx \]

### Step 2: Use substitution
Let \( u = \sqrt{2x + 1} \). Then, we differentiate \( u \):
\[ u^2 = 2x + 1 \implies 2x = u^2 - 1 \implies x = \frac{u^2 - 1}{2} \]
Now, we find \( dx \):
\[ dx = \frac{d}{du}\left(\frac{u^2 - 1}{2}\right) du = u \, du \]

### Step 3: Change the limits of integration
When \( x = 0 \):
\[ u = \sqrt{2(0) + 1} = \sqrt{1} = 1 \]
When \( x = 4 \):
\[ u = \sqrt{2(4) + 1} = \sqrt{9} = 3 \]
Thus, the limits change from \( x = 0 \) to \( x = 4 \) into \( u = 1 \) to \( u = 3 \).

### Step 4: Rewrite the integral
Substituting \( u \) and \( dx \) into the integral gives:
\[ I = \int_{1}^{3} 3^{u} \cdot u \, du \]

### Step 5: Integrate by parts
Let \( v = 3^u \) and \( dv = 3^u \ln(3) \, du \). We can use integration by parts:
\[ \int u \cdot 3^u \, du = u \cdot \frac{3^u}{\ln(3)} - \int \frac{3^u}{\ln(3)} \, du \]
\[ = u \cdot \frac{3^u}{\ln(3)} - \frac{3^u}{(\ln(3))^2} \]

### Step 6: Evaluate the integral
Now we evaluate from \( u = 1 \) to \( u = 3 \):
\[ I = \left[ u \cdot \frac{3^u}{\ln(3)} - \frac{3^u}{(\ln(3))^2} \right]_{1}^{3} \]

Calculating the upper limit \( u = 3 \):
\[ = 3 \cdot \frac{3^3}{\ln(3)} - \frac{3^3}{(\ln(3))^2} = 3 \cdot \frac{27}{\ln(3)} - \frac{27}{(\ln(3))^2} \]
Calculating the lower limit \( u = 1 \):
\[ = 1 \cdot \frac{3^1}{\ln(3)} - \frac{3^1}{(\ln(3))^2} = \frac{3}{\ln(3)} - \frac{3}{(\ln(3))^2} \]

### Step 7: Combine the results
Now, substituting back into the expression for \( I \):
\[ I = \left( \frac{81}{\ln(3)} - \frac{27}{(\ln(3))^2} \right) - \left( \frac{3}{\ln(3)} - \frac{3}{(\ln(3))^2} \right) \]
\[ = \frac{81 - 3}{\ln(3)} - \frac{27 - 3}{(\ln(3))^2} \]
\[ = \frac{78}{\ln(3)} - \frac{24}{(\ln(3))^2} \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{78}{\ln(3)} \]