1. We need to solve

$$\int_{\alpha}^{\alpha+1} \frac{dx}{(x+\alpha)(x+\alpha+1)}=\ln\left(\frac98\right).$$

2. Use partial fractions:

$$\frac{1}{(x+\alpha)(x+\alpha+1)} =\frac{1}{x+\alpha}-\frac{1}{x+\alpha+1}.$$

So the integral becomes

$$I=\int_{\alpha}^{\alpha+1}\left(\frac{1}{x+\alpha}-\frac{1}{x+\alpha+1}\right)dx.$$

3. Integrate:

$$I=\left[\ln|x+\alpha|-\ln|x+\alpha+1|\right]_{\alpha}^{\alpha+1}.$$

4. Substitute limits. At $x=\alpha+1$:

$$\ln|2\alpha+1|-\ln|2\alpha+2|.$$

At $x=\alpha$:

$$\ln|2\alpha|-\ln|2\alpha+1|.$$

Hence,

$$I=\ln|2\alpha+1|-\ln|2\alpha+2|-\ln|2\alpha|+\ln|2\alpha+1|.$$

So,

$$I=\ln\left(\frac{(2\alpha+1)^2}{(2\alpha)(2\alpha+2)}\right).$$

Since $(2\alpha)(2\alpha+2)=4\alpha(\alpha+1)$, we get

$$I=\ln\left(\frac{(2\alpha+1)^2}{4\alpha(\alpha+1)}\right).$$

5. Given that

$$I=\ln\left(\frac98\right),$$

therefore

$$\frac{(2\alpha+1)^2}{4\alpha(\alpha+1)}=\frac98.$$

Cross-multiply:

$$8(2\alpha+1)^2=9\cdot 4\alpha(\alpha+1).$$ $$2(2\alpha+1)^2=9\alpha(\alpha+1).$$

Expand:

$$2(4\alpha^2+4\alpha+1)=9\alpha^2+9\alpha$$ $$8\alpha^2+8\alpha+2=9\alpha^2+9\alpha$$ $$\alpha^2+\alpha-2=0.$$

Factor:

$$(\alpha+2)(\alpha-1)=0.$$

Thus,

$$\alpha=-2 \quad \text{or} \quad \alpha=1.$$

6. Now check validity of the integral. The integrand is

$$\frac{1}{(x+\alpha)(x+\alpha+1)}.$$

This is undefined at $x=-\alpha$ and $x=-\alpha-1$.

• If $\alpha=1$, interval is $[1,2]$, singularities are at $x=-1,-2$, which are outside the interval. Valid.
• If $\alpha=-2$, interval is $[-2,-1]$, singularities are at $x=2,1$, outside the interval. Valid.

So both values satisfy the equation.

7. Verify directly:

• For $\alpha=1$,

$$\int_1^2 \frac{dx}{(x+1)(x+2)} =\left[\ln\frac{x+1}{x+2}\right]_1^2 =\ln\frac{3/4}{2/3}=\ln\frac98.$$

• For $\alpha=-2$,

$$\int_{-2}^{-1} \frac{dx}{(x-2)(x-1)} =\left[\ln\frac{x-2}{x-1}\right]_{-2}^{-1} =\ln\frac{(-3)/(-2)}{(-4)/(-3)}=\ln\frac98.$$

Both work.

8. Therefore, from the given options, only $\boxed{-2}$ is present. So the correct option among the choices is B.