Integration by parts

Question

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?

Answer 1

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.