A value of `alpha` such that
`int_(alpha)^(alpha+1) (dx)/((x+a)(x+alpha+1))="loge"((9)/(8))` is

Options

A. 2
B. `-1/2`
C. `1/2`
D. `-2`

Explanation

To solve the problem, we need to find the value of \( \alpha \) such that:

\[ \int_{\alpha}^{\alpha+1} \frac{dx}{(x+\alpha)(x+\alpha+1)} = \log_e\left(\frac{9}{8}\right) \]

### Step 1: Simplify the integrand

We can simplify the integrand using partial fraction decomposition:

\[ \frac{1}{(x+\alpha)(x+\alpha+1)} = \frac{A}{x+\alpha} + \frac{B}{x+\alpha+1} \]

Multiplying through by the denominator:

\[ 1 = A(x+\alpha+1) + B(x+\alpha) \]

Expanding and rearranging gives:

\[ 1 = (A+B)x + (A\alpha + A + B\alpha) \]

Setting coefficients equal, we have:

1. \( A + B = 0 \)
2. \( A\alpha + A + B\alpha = 1 \)

From the first equation, we can express \( B \) in terms of \( A \):

\[ B = -A \]

Substituting into the second equation:

\[ A\alpha + A - A\alpha = 1 \implies A = 1 \]

Thus, \( B = -1 \). Therefore, we can rewrite the integrand as:

\[ \frac{1}{(x+\alpha)(x+\alpha+1)} = \frac{1}{x+\alpha} - \frac{1}{x+\alpha+1} \]

### Step 2: Integrate the simplified expression

Now we can integrate:

\[ \int_{\alpha}^{\alpha+1} \left(\frac{1}{x+\alpha} - \frac{1}{x+\alpha+1}\right) dx \]

This can be split into two separate integrals:

\[ \int_{\alpha}^{\alpha+1} \frac{1}{x+\alpha} dx - \int_{\alpha}^{\alpha+1} \frac{1}{x+\alpha+1} dx \]

Calculating each integral:

1. For the first integral:

\[ \int \frac{1}{x+\alpha} dx = \log_e(x+\alpha) \quad \text{evaluated from } \alpha \text{ to } \alpha+1 \]

This gives:

\[ \log_e(\alpha+1+\alpha) - \log_e(\alpha) = \log_e\left(\frac{2\alpha+1}{\alpha}\right) \]

2. For the second integral:

\[ \int \frac{1}{x+\alpha+1} dx = \log_e(x+\alpha+1) \quad \text{evaluated from } \alpha \text{ to } \alpha+1 \]

This gives:

\[ \log_e(\alpha+1+\alpha+1) - \log_e(\alpha+1) = \log_e\left(\frac{2\alpha+2}{\alpha+1}\right) \]

### Step 3: Combine the results

Combining both results, we have:

\[ \log_e\left(\frac{2\alpha+1}{\alpha}\right) - \log_e\left(\frac{2\alpha+2}{\alpha+1}\right) = \log_e\left(\frac{(2\alpha+1)(\alpha+1)}{2\alpha+2}\right) \]

Setting this equal to \( \log_e\left(\frac{9}{8}\right) \):

\[ \frac{(2\alpha+1)(\alpha+1)}{2\alpha+2} = \frac{9}{8} \]

### Step 4: Cross-multiply and simplify

Cross-multiplying gives:

\[ 8(2\alpha+1)(\alpha+1) = 9(2\alpha+2) \]

Expanding both sides:

\[ 16\alpha^2 + 24\alpha + 8 = 18\alpha + 18 \]

Rearranging gives:

\[ 16\alpha^2 + 6\alpha - 10 = 0 \]

### Step 5: Solve the quadratic equation

Dividing through by 2:

\[ 8\alpha^2 + 3\alpha - 5 = 0 \]

Using the quadratic formula:

\[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 8 \cdot (-5)}}{2 \cdot 8} \]

Calculating the discriminant:

\[ \sqrt{9 + 160} = \sqrt{169} = 13 \]

Thus:

\[ \alpha = \frac{-3 \pm 13}{16} \]

Calculating the two possible values:

1. \( \alpha = \frac{10}{16} = \frac{5}{8} \)
2. \( \alpha = \frac{-16}{16} = -1 \)

### Final Answer

The possible values of \( \alpha \) are \( \frac{5}{8} \) and \( -1 \).