A value of $$\alpha$$ such that $$\int_\alpha^{\alpha+1} \frac{dx}{(x + \alpha)(x + \alpha + 1)} = \log_e\left(\frac{9}{8}\right)$$ is

We start with the given condition

$$\int_\alpha^{\alpha+1}\frac{dx}{(x+\alpha)(x+\alpha+1)}=\log_e\!\left(\frac{9}{8}\right).$$

To simplify the integrand, we put $$t=x+\alpha.$$

Then when $$x=\alpha$$ we have $$t=\alpha+\alpha=2\alpha,$$ and when $$x=\alpha+1$$ we get $$t=(\alpha+1)+\alpha=2\alpha+1.$$

With this substitution $$dx=dt,$$ so the integral becomes

$$\int_{2\alpha}^{2\alpha+1}\frac{dt}{t(t+1)}.$$

Next we decompose the integrand using partial fractions. The standard identity

$$\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}$$

allows us to write

$$\int_{2\alpha}^{2\alpha+1}\left(\frac{1}{t}-\frac{1}{t+1}\right)dt.$$

Integrating term by term and recalling that $$\displaystyle\int\frac{dt}{t}=\ln|t|,$$ we obtain

$$\bigl[\ln|t|-\ln|t+1|\bigr]_{\,2\alpha}^{\,2\alpha+1}.$$

Applying the limits gives

$$\Bigl(\ln(2\alpha+1)-\ln(2\alpha+2)\Bigr)-\Bigl(\ln(2\alpha)-\ln(2\alpha+1)\Bigr).$$

Collecting like terms,

$$2\ln(2\alpha+1)-\ln(2\alpha)-\ln(2\alpha+2).$$

This must equal the given logarithm, so

$$2\ln(2\alpha+1)-\ln(2\alpha)-\ln(2\alpha+2)=\ln\!\left(\frac{9}{8}\right).$$

Using the property $$\ln a-\ln b=\ln\!\left(\frac{a}{b}\right)$$ twice, we combine the left-hand side into a single logarithm:

$$\ln\!\left(\frac{(2\alpha+1)^2}{2\alpha(2\alpha+2)}\right)=\ln\!\left(\frac{9}{8}\right).$$

Exponentiating both sides removes the logarithms:

$$\frac{(2\alpha+1)^2}{2\alpha(2\alpha+2)}=\frac{9}{8}.$$

Simplifying the denominator $$2\alpha(2\alpha+2)=4\alpha(\alpha+1),$$ we have

$$\frac{(2\alpha+1)^2}{4\alpha(\alpha+1)}=\frac{9}{8}.$$

Cross-multiplying gives

$$8(2\alpha+1)^2=9\cdot4\alpha(\alpha+1).$$

Dividing both sides by 4,

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

Expanding the square $$\bigl(2\alpha+1\bigr)^2=4\alpha^2+4\alpha+1,$$ we get

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

This simplifies to

$$8\alpha^2+8\alpha+2=9\alpha^2+9\alpha.$$

Bringing all terms to one side,

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

We now solve the quadratic $$\alpha^2+\alpha-2=0.$$

The quadratic formula $$\alpha=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ with $$a=1,\;b=1,\;c=-2$$ gives

$$\alpha=\frac{-1\pm\sqrt{1^2-4(1)(-2)}}{2}=\frac{-1\pm\sqrt{9}}{2}=\frac{-1\pm3}{2}.$$

Thus $$\alpha_1=\frac{-1+3}{2}=1,\qquad \alpha_2=\frac{-1-3}{2}=-2.$$

Both values keep the integrand finite over the interval, but only $$\alpha=-2$$ appears among the four offered choices.

Hence, the correct answer is Option C.