I'm trying to answer this integral
$$ \int \frac{dx}{x\sqrt{x^2+2x-1}} $$
using the Reciprocal substitution method. How should I approach this?
Should I first rewrite or simplify the problem before I proceed to answer? if yes, someone please help me, an answer on this problem and as well an explanation would be so much appreciated too.
Thanks and advance and have a nice day.
Problem $$\int \:\frac{1}{x\sqrt{x^2+2x-1}}dx$$ Complete the square $$\int \:\frac{1}{x\sqrt{\left(x+1\right)^2-2}}dx$$ Substitute $$u=x+1\rightarrow \:\frac{du}{dx}=1\rightarrow \:du=dx$$ and $$u=x+1\rightarrow \:x=u-1\rightarrow \:\:\frac{1}{x}=\frac{1}{u-1}$$ will becomes $$\int \:\frac{1}{\left(u-1\right)\sqrt{u^2-2}}du$$ Perform trigonometric substitution $$u=\sqrt{2}sec\left(v\right)\rightarrow v=arcsec\left(\frac{u}{\sqrt{2}}\right),du=\sqrt{2}sec\left(v\right)tan\left(v\right)dv$$ will becomes $$\int \:\frac{\sqrt{2}sec\left(v\right)tan\left(v\right)}{\left(\sqrt{2}sec\left(v\right)−1\right)\sqrt{2sec^2\left(v\right)−2}}dv$$ Simplify by using $$2sec^2\left(v\right)−2=2tan^2\left(v\right)$$ and becomes $$\int \:\frac{sec\left(v\right)}{\sqrt{2}sec\left(v\right)−1}dv$$ $$\frac{1}{\sqrt{2}}\int \:\frac{1}{\sqrt{2}sec\left(v\right)-1}dv+\frac{1}{2}\int \:dv$$ So, i think that you can take from here. For references, the final answer should be: $$\frac{2arctan\left(\frac{\sqrt{\sqrt{2}+1}\sqrt{x-\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}\sqrt{x+\sqrt{2}+1\:}}\right)}{\sqrt{\sqrt{2}+1}\sqrt{\sqrt{2}-1}}+c$$