A) \[\frac{1}{2}\]
B) \[2\]
C) \[-\frac{1}{2}\]
D) \[-2\]
Correct Answer: D
Solution :
\[\int\limits_{\alpha }^{\alpha +1}{\frac{\left( x+\alpha +1 \right)-\left( x+\alpha \right)}{\left( x+\alpha \right)\left( x+\alpha +1 \right)}dx}=\] \[\left( \ell n\left| x+\alpha \right|-\ell n\left| x+\alpha +1 \right| \right)_{\alpha }^{\alpha +1}\] \[=\ell n\left| \frac{2\alpha +1}{2\alpha +2}\times \frac{2\alpha +1}{2\alpha } \right|=\ell n\frac{9}{8}\]\[\Rightarrow \alpha =-2,1\]
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