Question

If $f\left(x\right)=\underset{n\to \infty }{lim}(2x+4x^3+\dots +2^n{x}^{2n-1})$, where $x\in (0,\frac{1}{\sqrt{2}}),$ then $\int f\left(x\right)dx$ is equal to

• A. $\log \left(\frac{1}{\sqrt{1+2x^2}}\right)+C$
• B. $\log \sqrt{1-2x^2}+C$
• C. $\log \left(\frac{1}{\sqrt{1-2x^2}}\right)+C$
• D. $\log (1-2x^2)+C$

Answer 1

The correct option is C $\log \left(\frac{1}{\sqrt{1-2x^2}}\right)+C$

We have,
$f\left(x\right)=\underset{n\to \infty }{lim}(2x+4x^3+\dots +2^n{x}^{2n-1})$

Let $S=2x+4x^3+...+2^n{x}^{2n-1}$
This is a G.P. whose first term is $2x$ and common ratio is $2x^2$, then
$S=\frac{2x[1-(2x^2{)}^n]}{(1-2x^2)}$
We know that
$x\in (0,\frac{1}{\sqrt{2}})\Rightarrow 2x^2\in (0,1)$

Now,
$f\left(x\right)=\underset{n\to \infty }{lim}\frac{2x[1-(2x^2{)}^n]}{(1-2x^2)}\Rightarrow f\left(x\right)=\frac{2x}{1-2x^2}$

Therefore,
$I=\int \frac{2x}{1-2x^2}dx$
Assuming $1-2x^2=t\Rightarrow -4xdx=dt$
$\Rightarrow I=-\frac{1}{2}\int \frac{dt}{t}\Rightarrow I=-\frac{1}{2}\log |t|+C\Rightarrow I=-\frac{1}{2}\log |1-2x^2|+C$

As $2x^2\in (0,1)\Rightarrow 1-2x^2\in (0,1)$, so
$I=\log \left(\frac{1}{\sqrt{1-2x^2}}\right)+C$