Definite Integration Question 238
Question: The value of integral $ \int_0^{1}{\frac{x^{b}-1}{\log x}}dx $ is
Options:
A) $ \log b $
B) $ 2\log (b+1) $
C) $ 3\log b $
D) None of these
Show Answer
Answer:
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}} $
therefore $ I(b)=\int{\frac{db}{b+1}+c=\log (b+1)+c} $
If $ b=0 $ , then $ I(b)=0 $ , so $ c=0 $
therefore $ I(b)=\log (b+1) $ .
BETA
