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.

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

##### 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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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

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