A value of such that _^+1(dx)/((x+)(x++1))=_e98 is:

Question

A value of $\alpha$ such that $\displaystyle\int_\alpha^{\alpha+1}\frac{dx}{(x+\alpha)(x+\alpha+1)}=\log_e\dfrac98$ is:

Accepted Answer

Using partial fractions $\dfrac1{(x+\alpha)(x+\alpha+1)}=\dfrac1{x+\alpha}-\dfrac1{x+\alpha+1}$:
$$\int_\alpha^{\alpha+1}\left(\frac1{x+\alpha}-\frac1{x+\alpha+1}\right)dx=\left[\ln\frac{x+\alpha}{x+\alpha+1}\right]_\alpha^{\alpha+1}=\ln\frac{2\alpha+1}{2\alpha+2}\cdot\frac{2\alpha+2}{2\alpha+1}\cdots$$

Carefully: $=\ln\dfrac{(2\alpha+1)^2}{2\alpha(2\alpha+2)}=\ln\dfrac98$ (given)

$$\Rightarrow\frac{(2\alpha+1)^2}{\alpha(\alpha+1)}=\frac92\Rightarrow8\alpha^2+8\alpha+2=9\alpha^2+9\alpha\Rightarrow\alpha^2+\alpha-2=0\Rightarrow\alpha=1,-2$$

The listed value is $\alpha=-2$.

Solutions · 1