Question

If $f\left(x\right)=\underset{n\to \infty }{lim}[2x+4x^3+6x^5+.......+2b{x}^{2n-1}]$ $(0<x<1)$ then $\int f\left(x\right)dx$ is equal to

• A. $-\sqrt{1-x^2}+c$
• B. $\frac{1}{\sqrt{1-x^2}}+c$
• C. $\frac{1}{-x^2-1}+c$
• D. $\frac{1}{1-x^2}+c$

Answer 1

The correct option is D $\frac{1}{1-x^2}+c$

$\begin{array}{c}f\left(x\right)={lim}_{x\to \infty }\left[2x+4x^3+6x^5+........2+x^{\left(2x-1\right)}\right]\\ ={\sum }_{n=1}^{\propto }2n{x}^{2n-1}0<x<1\\ \Rightarrow then,\int fxdx={\sum }_{n=1}^{\propto }2n\frac{x^{2n}}{2n}+C_1=(1+\\ {\sum }_{n=1}^{\propto }{x}^{2n}+c)=⟨{\sum }_{n=0}^{\propto }{x}^{2n}+c={\sum }_{n=0}^{\propto }{\left(x^2\right)}^n+c⟩=\frac{1}{\left(1-x^2\right)}+c\end{array}$