Question

$\int \frac{dx}{(\sqrt{1+x^2}-x{)}^n}(n\ne \pm 1)=\frac{1}{2}(\frac{z^{n+1}}{n+1}+\frac{z^{n-1}}{n-1})+O$
where

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

Answer 1

The correct option is C $z=x+\sqrt{1+x^2}$

Given:
$\int \frac{dx}{{(\sqrt{1+x^2}-x)}^n}=\frac{1}{2}[\frac{z^{n+1}}{(n+1)}+\frac{z^{n-1}}{(n-1)}]+C$
solution;
$t=(\sqrt{1+x^2}-x)t+x=\sqrt{1+x^2}{t}^2+x^2+2xt=1+x^{2}x=\left(\frac{1-t^2}{2t}\right)=\frac{1}{2t}-\frac{t}{2}\frac{dx}{dt}=\frac{1}{2t^2}-\frac{1}{2}=-\frac{1}{2}\left(\frac{1+t^2}{t^2}\right)\therefore I=\int (-\frac{1}{2})\times \left(\frac{1+t^2}{t^2}\right)\frac{1}{t^n}dt=-\frac{1}{2}[\int t^{-(2+n)}dx+\int t^{-n}dt]=\frac{1}{2}[\frac{t^{-(1+n)}}{-(1+n)}+\frac{t^{-(n-1)}}{-(n-1)}]=z=t^{-1}=\sqrt{1+x^2}+x$
Hence the correct ans is $\sqrt{1+x^2}+x$