Evaluate :
(i) Find the value a such `int_(0)^(a)(1)/(e^(x)+4e^(x)+5)dx = ln 3sqrt(2)`.

Answer

To evaluate the integral and find the value of \( a \) such that

\[ \int_{0}^{a} \frac{1}{e^{x} + 4e^{x} + 5} \, dx = \ln(3\sqrt{2}), \]

we will follow these steps:

### Step 1: Simplify the Denominator
The denominator can be rewritten as:

\[ e^{x} + 4e^{x} + 5 = 5e^{x} + 5 = 5(e^{x} + 1). \]

### Step 2: Rewrite the Integral
Now, the integral becomes:

\[ \int_{0}^{a} \frac{1}{5(e^{x} + 1)} \, dx = \frac{1}{5} \int_{0}^{a} \frac{1}{e^{x} + 1} \, dx. \]

### Step 3: Evaluate the Integral
The integral \( \int \frac{1}{e^{x} + 1} \, dx \) can be computed using the substitution \( u = e^{x} \), which gives \( du = e^{x} \, dx \) or \( dx = \frac{du}{u} \). The limits change accordingly from \( x = 0 \) to \( x = a \):

- When \( x = 0 \), \( u = e^{0} = 1 \).
- When \( x = a \), \( u = e^{a} \).

Thus, we have:

\[ \int_{0}^{a} \frac{1}{e^{x} + 1} \, dx = \int_{1}^{e^{a}} \frac{1}{u + 1} \cdot \frac{du}{u} = \int_{1}^{e^{a}} \frac{1}{u(u + 1)} \, du. \]

### Step 4: Partial Fraction Decomposition
We can decompose \( \frac{1}{u(u + 1)} \) as follows:

\[ \frac{1}{u(u + 1)} = \frac{A}{u} + \frac{B}{u + 1}. \]

Multiplying through by \( u(u + 1) \) gives:

\[ 1 = A(u + 1) + Bu. \]

Setting \( u = 0 \) gives \( A = 1 \). Setting \( u = -1 \) gives \( B = -1 \). Thus:

\[ \frac{1}{u(u + 1)} = \frac{1}{u} - \frac{1}{u + 1}. \]

### Step 5: Integrate
Now we can integrate:

\[ \int_{1}^{e^{a}} \left( \frac{1}{u} - \frac{1}{u + 1} \right) \, du = \left[ \ln |u| - \ln |u + 1| \right]_{1}^{e^{a}}. \]

Calculating this gives:

\[ \left( \ln(e^{a}) - \ln(e^{a} + 1) \right) - \left( \ln(1) - \ln(2) \right) = a - \ln(e^{a} + 1) + \ln(2). \]

### Step 6: Substitute Back
Now substituting back into our integral:

\[ \frac{1}{5} \left( a - \ln(e^{a} + 1) + \ln(2) \right) = \ln(3\sqrt{2}). \]

### Step 7: Solve for \( a \)
Multiplying both sides by 5 gives:

\[ a - \ln(e^{a} + 1) + \ln(2) = 5\ln(3\sqrt{2}). \]

This simplifies to:

\[ a - \ln(e^{a} + 1) = 5\ln(3\sqrt{2}) - \ln(2). \]

### Step 8: Final Rearrangement
Rearranging gives:

\[ a - \ln(e^{a} + 1) = 5\ln(3) + 5\ln(\sqrt{2}) - \ln(2) = 5\ln(3) + \frac{5}{2}\ln(2) - \ln(2) = 5\ln(3) + 2\ln(2). \]

### Step 9: Conclusion
Now we can solve this equation numerically or graphically to find the value of \( a \).