Question

How do you evaluate the integral `int 1/(xsqrt(2+x^2))`?

Answer 1

`1/sqrt2lnabs(x/(sqrt(x^2+2)+sqrt2))+C`

Explanation:

`I=intdx/(xsqrt(2+x^2))`

Use the substitution `x=sqrt2tantheta`. This implies that `x^2=2tan^2theta` and `dx=sqrt2sec^2thetad theta`.

Plugging this in shows that:

`I=int(sqrt2sec^2thetad theta)/(sqrt2tanthetasqrt(2+2tan^2theta))`

Note that `sqrt(2+2tan^2theta)=sqrt2sqrt(1+tan^2theta)=sqrt2sectheta`. This comes from the identity `1+tan^2theta=sec^2theta`.

`I=int(sqrt2sec^2thetad theta)/(sqrt2tantheta(sqrt2sectheta))=1/sqrt2intsectheta/tanthetad theta`

Rewriting with `sintheta` and `costheta`:

`I=1/sqrt2int1/costhetacostheta/sinthetad theta=1/sqrt2intcscthetad theta=-1/sqrt2lnabs(csctheta+cottheta)`

From the original substitution we have `tantheta=x/sqrt2`. This correlates to a right triangle with the side opposite `theta` being `x` and the side adjacent to `theta` being `sqrt2`. The Pythagorean theorem gives the hypotenuse as `sqrt(x^2+2)`.

Thus, `csctheta` is the hypotenuse over the opposite side, or `csctheta=sqrt(x^2+2)/x` and `cottheta=1/tantheta=sqrt2/x`.

So:

`I=-1/sqrt2lnabs((sqrt(x^2+2)+sqrt2)/x)+C`

We can rewrite this using `Bln(A)=ln(A^B)`. In this case, we will use `-ln(A)=ln(A^-1)=ln(1/A)`.

`I=1/sqrt2lnabs(x/(sqrt(x^2+2)+sqrt2))+C`