A) 7
B) 8
C) 9
D) 10
Correct Answer: C
Solution :
Let\[I=\int_{0}^{1}{\frac{({{x}^{\alpha }}-1)dx}{\log x}}\] \[\frac{dI}{d\alpha }=\int_{0}^{1}{\frac{{{x}^{\alpha }}\log x}{\log x}}dx\] \[\Rightarrow \] \[\frac{dI}{d\alpha }=\int_{0}^{1}{{{x}^{\alpha }}dx}\] \[=\left[ \frac{{{x}^{\alpha +1}}}{\alpha +1} \right]_{0}^{1}\] \[\Rightarrow \] \[\frac{dI}{d\alpha }=\frac{1}{\alpha +1}\] \[\Rightarrow \] \[dI=\frac{1}{\alpha +1}d\alpha \] On integrating, we get \[I=\log (\alpha +1)+c\] Now, if\[\alpha =0,\]then \[I=\int_{0}^{1}{\frac{({{x}^{0}}-1)}{\log x}}dx\] \[\Rightarrow \] \[I=\int_{0}^{1}{0\,dx}=0\] \[\therefore \]From Eq. (i), \[0=\log 1+c\] \[\Rightarrow \] \[c=0\] Hence, \[I=log(\alpha +1)\]
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