Here is the question once again:
Deduce the integral $\int{\sqrt{x^2+1}} \ \mathrm{d}x$ using integration by parts, given the following identity: $$\int{\frac{1}{\sqrt{x^2+1}} \ \mathrm{d}}x=\ln(x+\sqrt{x^2+1})+k$$
I know that the integral can be solved by using trigonometric substitutions, but I am looking for another way as that's not what is asked in the question.
Here is what I have done so far:
$$\int{\sqrt{x^2+1}} \ \mathrm{d}x=x \cdot \sqrt{x^2+1}-\underbrace{\int{\frac{x^2}{\sqrt{x^2+1}} \ \mathrm{d}x}}_{\text{$I_1$}}$$
$$I_1=x^2 \cdot \ln{(x+\sqrt{x^2+1})}-\underbrace{\int{\ln{(x+\sqrt{x^2+1})}\cdot 2x \ \mathrm{d}x}}_{\text{$I_2$}}$$
Solving $I_2$ using integration by parts gives an expression which includes $I_1$ so I got into a loop. I'm out of ideas so I would be glad if you could help me out, thank you for your time.
Notice that
$$\int \sqrt {x^2 + 1} \ \Bbb d x = \int (x')\sqrt {x^2 + 1} \ \Bbb d x = x \sqrt {x^2 + 1} - \int x \frac {x} {\sqrt {x^2 + 1}} \ \Bbb d x = \\ x \sqrt {x^2 + 1} - \int \frac {x^2 + 1 -1} {\sqrt {x^2 + 1}} \ \Bbb d x = x \sqrt {x^2 + 1} - \int \sqrt {x^2 + 1} - \frac 1 {\sqrt {x^2 + 1}} \ \Bbb d x = \\ - \int \sqrt {x^2 + 1} \ \Bbb d x + x \sqrt {x^2 + 1} + \ln (x + \sqrt {x^2 + 1})$$
whence it follows that
$$2 \int \sqrt {x^2 + 1} \ \Bbb d x = x \sqrt {x^2 + 1} + \ln (x + \sqrt {x^2 + 1}) $$
so that
$$\int \sqrt {x^2 + 1} \ \Bbb d x = \frac 1 2 x \sqrt {x^2 + 1} + \frac 1 2 \ln (x + \sqrt {x^2 + 1}) + C$$
where $C \in \Bbb R$ is an arbitrary integration constant added at the end of the calculations.