MCQ
The value of integral $\int_0^1 {\frac{{{x^b} - 1}}{{\log x}}} \,dx$ is
- A$\log b$
- B$2\log (b + 1)$
- C$3\log b$
- ✓None of these
✓
Answer
Correct option: D.
None of these
d
(d) Let $I(b) = \int_0^1 {\frac{{{x^b} - 1}}{{\log x}}} dx $
(d) 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}}} $
==> $I(b) = \int {\frac{{db}}{{b + 1}} + c = \log (b + 1) + c} $
If $b = 0$, then $I(b) = 0$,
so $c = 0$==>$I(b) = \log (b + 1)$.
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