Mathematics

If $$ \displaystyle f(x)=\lim_{n\rightarrow \infty }(2x+4x^{3}+......+2^{n}x^{2n-1})\left ( 0<x<\frac{1}{\sqrt{2}} \right )$$, then the value of $$\displaystyle\int f(x) dx$$ is equal to
$$\textbf{Note}$$: $$c$$ is the constant of integration.


ANSWER

$$ \displaystyle \log\left ( \frac{1}{\sqrt{1-2x^{2}}} \right )+c$$


SOLUTION

When $$ n \rightarrow \infty$$, $$f(x)$$ becomes an infinites series of G.P. with $$a=2x$$, $$r=2x^2$$

Hence, $$f(x)=\dfrac{2x}{1-2x^2}$$

Since, $$0<x<\dfrac1{\sqrt2}$$

So, the integral $$I = \displaystyle\int f(x)\>dx$$ becomes $$ \displaystyle \int \frac{2x}{1-2x^{2}}dx$$

Substitute, $$1-2x^{2}=t\Rightarrow -4x\>dx=dt$$

After substitution, the integral becomes

$$\displaystyle -\frac{1}{2}\int \frac{dt}{t}=-\frac{1}{2}\log(t)+c$$

$$ \displaystyle =-\frac{1}{2}\log (1-2x^2) +c$$

$$ \displaystyle I =\log\left ( \frac{1}{\sqrt{1-2x^{2}}} \right )+c$$


Hence, option C.

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