A value of α such that ∫ _α ^α +1 dx (x+α )(x+α +1)=log_e( 9 8) is

$$I=\int _{\alpha }^{\alpha +1}\frac{dx}{(x+\alpha )(x+\alpha +1)}=\int _{\alpha }^{\alpha +1}(\frac{1}{x+\alpha }–\frac{1}{x+\alpha +1 })dx$$I=∫αα+1​(x+α)(x+α+1)dx​=∫αα+1​(x+α1​–x+α+11​)dx

Using partial fractions

$$={[ln|x+\alpha |-ln|x+\alpha +1|]}_{\alpha }^{\alpha +1}$$=[ln∣x+α∣−ln∣x+α+1∣]αα+1​

$$={[ln|\frac{x+\alpha }{x+\alpha +1}|]}_{\alpha }^{\alpha +1}$$=[ln∣x+α+1x+α​∣]αα+1​

$$=ln|\frac{2\alpha +1}{2\alpha +2}|-ln|\frac{2\alpha }{2\alpha +1}|$$=ln∣2α+22α+1​∣−ln∣2α+12α​∣

$$=ln|\frac{{(2\alpha +1)}^{2}}{{(2\alpha +1)}^{2}-1}|=\mathrm{ln}⁡\frac{9}{8}$$=ln∣(2α+1)2−1(2α+1)2​∣=ln⁡89​

$$\Rightarrow {(2\alpha +1)}^{2}=9$$⇒(2α+1)2=9

$$\Rightarrow 2\alpha +1=\pm 3$$⇒2α+1=±3

$\Rightarrow \alpha =1$⇒α=1 or $-2$−2.