A) \[\log b\]
B) \[2\log (b+1)\]
C) \[3\log b\]
D) None of these
Correct Answer: D
Solution :
Let \[I(b)=\int_{0}^{1}{\frac{{{x}^{b}}-1}{\log x}}dx\Rightarrow I'(b)=\int_{0}^{1}{\frac{{{x}^{b}}\log x}{\log x}dx}\] (If \[I(\alpha )=\int_{0}^{b}{f(x,\alpha )dx}\], then \[I'(\alpha )=\int_{0}^{b}{f'(x,\alpha )dx}\], where \[f'(x,\alpha )\] is derivative of \[f(x,\alpha )\] w.r.t. \[\alpha \] keeping x constant) \[{I}'(b)=\int_{0}^{1}{{{x}^{b}}dx=\frac{1}{b+1}}\] Þ \[I(b)=\int{\frac{db}{b+1}+c=\log (b+1)+c}\] If \[b=0\], then \[I(b)=0\], so \[c=0\]Þ\[I(b)=\log (b+1)\].
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