For 0 f(x)= _ n (1+x) (1+x^ 2 ) (1+x^ 4 ) (1+x^ 2^ n ) then f(x) 1-x log e xdx equals
Mathematics · JEE Advanced · NTA Exams — Integral Calculus
For 0 < x < 1, let
\(f(x)=\lim _{n \rightarrow \infty}(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2^{n}}\right)\)
then \(\int \frac{f(x)}{1-x}\) loge xdx equals
\(f(x)=\lim _{n \rightarrow \infty}(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2^{n}}\right)\)
then \(\int \frac{f(x)}{1-x}\) loge xdx equals
- \(\log _{e}\left(\frac{x}{1-x}\right)+c\)
- \(-\log _{e}\left(\frac{x}{1-x}\right)+\frac{\log _{e} x}{1-x}+c\)
- \(\frac{\log _{e} x}{1-x}+\log _{e}(1-x)+c\)
- x loge x + loge (1 – x) + c
Answer
(B) - _ e ( x 1-x )+ _ e x 1-x +c
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