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Question 2)b) part (ii) is the section that I'm having trouble with:


I don't understand the method used in the solutions; how would you deduce the first line or is that something you should know?


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There's a reason why LaTeX formatting is used on the site - it's a lot easier to understand than going through someone else's handwriting, parts of which are probably irrelevant. – nbubis Feb 3 '13 at 15:47
I understand but I didn't think this was that hard to read and I doubt any of it is irrelevant, considering it's the official solution from the mark scheme. – Mathlete Feb 3 '13 at 16:24
up vote 1 down vote accepted

Observe that $$\frac{d\ln(x+\sqrt{x^2+1})}{dx}=\frac1{x+\sqrt{x^2+1}}\left(1+\frac{2x}{2\sqrt{x^2+1}}\right)=\frac1{\sqrt{x^2+1}}$$

Now using Integration by parts formula: $\int uvdx=u\int v dx-\left(\frac{du}{dx}\int vdx\right)dx$,

$$\int \ln(x+\sqrt{x^2+1})\cdot 1 dx$$ $$=\ln(x+\sqrt{x^2+1})\int dx-\left(\frac{d\ln(x+\sqrt{x^2+1})}{dx}\int dx\right)dx$$ $$=x\ln(x+\sqrt{x^2+1})-\int \frac{x}{\sqrt{x^2+1}} dx$$

Now, put $x^2+1=y^2$ in the last integral

So, $ xdx=ydy$ $$\int \frac{x}{\sqrt{x^2+1}} dx=\int\frac{ydy}y=y+c=\sqrt{x^2+1}+c$$ where $c$ is the arbitrary constant for indefinite integral.

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Sorry, how did you get to the second line? And is that dx supposed to be there? – Mathlete Feb 3 '13 at 16:23
Thanks, I'm familiar with the method but a few more brackets wouldn't have gone amiss. I can see it now. – Mathlete Feb 3 '13 at 16:26
@Mathlete, please find the edited answer. – lab bhattacharjee Feb 3 '13 at 16:29

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