JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Medium

1 of 1

Analyzing +4 | -1

For

0 < x < 1

, let

f(x)=\lim \limits_{{n \rightarrow \infty}}(1+x)\left(1+x^2\right)\left(1+x^4\right) \cdots\left(1+x^{2^n}\right)

then

\int \frac{f(x)}{1-x} \log _e x d x

equals

View answer and explanation

Correct option: B

-\log _e\left(\frac{x}{1-x}\right)+\frac{\log _e x}{1-x}+c

Option B is correct because the integral can be solved by simplifying the limit, then using integration by parts.

More Questions from INDEFINITE INTEGRATION

Practice All Chapter MCQs →

\int \sin ^5 x \cos ^4 x d x=

f(x)

\int\limits_{0}^{t^2} x f(x) d x=\frac{2}{5} t^5

f\left(\frac{4}{25}\right)

\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{\sin ^7 x}+C

\int {e}^{{x}^2}\left(2-\frac{1}{{x}^2}\right) {de}={e}^{{x}^2} {f}({x})+{C}

{f}\left(\frac{1}{2}\right)=2

f(1)

{C}

\int \frac{x^7}{\left(1-x^2\right)^5} d x

\int \,\frac{{a{x^3} + \,\,b{x^2} + c}}{{{x^4}}}\;\,dx