Calculate $\int\frac{dx}{x\sqrt{x^2-2}}$.

The exercise is:

Calculate:$$\int\frac{dx}{x\sqrt{x^2-2}}$$

My first approach was:

Let $z:=\sqrt{x^2-2}$ then $dx = dz \frac{\sqrt{x^2-2}}{x}$ and $x^2=z^2+2$ $$\int\frac{dx}{x\sqrt{x^2-2}} = \int\frac{1}{x^2}dz = \int\frac{1}{z^2+2}dz = \frac{1}{2}\int\frac{1}{(\frac{z}{\sqrt{2}})^2 + 1}dz = \frac{1}{2} \arctan(\frac{z}{\sqrt{2}})$$

$$= \frac{1}{2}\arctan(\frac{\sqrt{x^2-2}}{\sqrt{2}})$$

Yet when e.g. Maple derives the expression, it turns out my solution is false. I also cannot find the mistake, and I'd be glad if someone could point it out to me.

I am interested in your solutions, but in another approach I substituted $z:=\frac{\sqrt{2}}{x}$ and multiplied the integrand with $\frac{\sqrt{2}x}{\sqrt{2}x}$ and it worked out just fine.

• It should be $\frac{1}{\sqrt{2}}\arctan(\frac{\sqrt{x^2-2}}{\sqrt{2}})$. – Mhenni Benghorbal Aug 12 '14 at 22:23
• @MhenniBenghorbal I've noticed. But how's there a mistake in the third equality? – Nhat Aug 12 '14 at 22:26

In the step$$\frac{1}{2}\int\frac{1}{(\frac{z}{\sqrt{2}})^2 + 1}dz=\frac{1}{2} \arctan(\frac{z}{\sqrt{2}})$$ you made a small mistake. The integral $$\int\frac{1}{\left(\frac{x}{a}\right)^2+1}dx=a\cdot \arctan\left(\frac{x}{a}\right)$$ not $\arctan\left(\frac{x}{a}\right)$
So if you multiply by the factor of $\sqrt{2}$ that you forgot, your answer is right.
If you replace $x=\sqrt{2} \cosh t$, you end with: $$\begin{eqnarray*} I = \frac{1}{\sqrt{2}}\int \frac{dt}{\cosh t}=\sqrt{2}\arctan\left(\tanh\frac{t}{2}\right)&=&\sqrt{2}\arctan\left(\frac{\sqrt{x^2-2}}{x+\sqrt{2}}\right)\\&=&\frac{1}{\sqrt{2}}\arctan\sqrt{\frac{x^2-2}{2}}.\end{eqnarray*}$$ Your solution is right, except for the $\sqrt{2}$ factor.
The mistake is that you are integrating over $z$,you can do a substitution $$u=\frac{z}{\sqrt{2}},dz=\sqrt{2}du\\\frac{1}{2}\int\frac{\sqrt{2}du}{u^2+1}=\frac{\sqrt{2}}{2}\arctan(u)=\frac{1}{\sqrt{2}}\arctan(\frac{\sqrt{x^2-2}}{\sqrt{2}})$$