Question

If $f:R\to R,f\left(x\right)=\frac{x}{\sqrt{1+x^2}}\&g\left(x\right)=(fofofo\dots 2018\mathrm{times}(x\left)\right)$ , find $\int g\left(x\right)dx$.

Answer 1

$f:R\to R$

$f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$

$g\left(x\right)=fofofo....f\left(x\right)$ (2018 times)

$fof\left(x\right)=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}=\frac{x}{\sqrt{1+2x^2}}$

$fofof\left(x\right)=\frac{\frac{x}{\sqrt{1+{2x}^2}}}{\sqrt{1+\frac{x^2}{1+2x^2}}}=\frac{x}{\sqrt{1+3x^2}}$

$g\left(x\right)=fofofo....f\left(x\right)=\frac{x}{\sqrt{1+2018x^2}}$

$\int g\left(x\right)dx=\int \frac{xdx}{\sqrt{1+2018}{x}^2}$

$\frac{1}{\sqrt{2018}}\int \frac{xdx}{\sqrt{x^2+\frac{1}{\sqrt{2018}}}}$

$=\frac{1}{\sqrt{2018}}\sqrt{x^2+\frac{1}{\sqrt{2018}}}=\frac{\sqrt{2018x^2+1}}{2018}$