How to compute this integral?

Question

I don't really know where to start with this.

$$\int \sqrt{x^2+y^2+1}\quad dx$$

Answer 1

Take $x = \sqrt{y^2+1} \sinh u$. This yields

$$ \left (y^2+1 \right )\int \cosh^2 u du $$

which is more straight-forward to compute.

Answer 2

Hint: Let $D=y^2+1$, because y is an independent variable, then use trignometric substitution.

Answer 3

Well, you may easily solve this one without any substitutions - just apply inegration by parts. Take $~~u = \sqrt{1+y^2+x^2}~~$ and $~~ dv = dx ~~$. Then you'll get that

$I = \int \sqrt{1+y^2+x^2} ~~ dx = x \sqrt{1 + y^2 + x^2} - \int {\frac{x^2}{1+y^2+x^2}} ~~ dx = x \sqrt{1 + y^2 + x^2} - \int {(\sqrt{1+y^2+x^2} - \frac{1}{1+\frac{x^2}{1+y^2}})} ~~ dx = x \sqrt{1 + y^2 + x^2} - I + \int{\frac{1}{1+\frac{x^2}{1+y^2}})} $

The last one is a standard integral. All you need is to "extract" I.