Q. If $f(x) = \lim_{n \rightarrow \infty}{\left\lbrack 2x + 4x^{3} + ... + 2nx^{2n - 1} \right\rbrack(0 < x < 1)}$, then $\int_{}^{}{f(x)dx}$ is equal to
A) $- \sqrt{1 - x^{2}}$
B) $\frac{1}{\sqrt{1 - x^{2}}}$
C) $\frac{1}{x^{2} - 1}$
D) $\frac{1}{1 - x^{2}}$
Explanation
(d)
Let $g_{n}(x) = 1 + x^{2} + x^{4} + ... + x^{2n} = \frac{x^{2n + 2} - 1}{x^{2} - 1}$
$\therefore h_{n}(x) = g_{n}^{'}(x) = \frac{2x({nx}^{2n + 2} - (n + 1)x^{2n} + 1)}{\left( x^{2} - 1 \right)^{2}}$
Now, $f(x) = \lim_{n \rightarrow \infty}{h_{n}(x) = \frac{2x}{\left( x^{2} - 1 \right)^{2}}}$
As $0 < x < 1$
Thus, $\int_{}^{}{f(x)dx = \int_{}^{}{\frac{2x}{\left( x^{2} - 1 \right)^{2}}dx}}$
$= - \frac{1}{x^{2} - 1} = \frac{1}{1 - x^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.