# How to integrate this fraction: $\int\frac{1}{1-2x^2}dx$?

I'm not sure how to integrate this: $$\int\frac{1}{1-2x^2}dx$$

I think it has to be this: $$-2\cdot \arctan(x)$$

Or this: $$\arctan(\sqrt{-2x^2})$$

• Have you tried partial fractions? Commented Nov 7, 2015 at 20:09
• Please, don't use $z$ for a real integration, it can be confusing with a complex integral.
– Surb
Commented Nov 7, 2015 at 20:10
• @Surb Maybe the $z$ comes from a substitution. Commented Nov 7, 2015 at 20:14
• You can write it as an arctangent of $x\cdot\sqrt{-2}$ (times an appropriate constant), and then have an arctangent of an imaginary number of $x$ is real, but then you need to look at the multiple-valued nature of the arctangent. You can also use partial fractions as suggested in some answers below and get a logarithm that has only real values when $x$ is real, and there you see how the logarithm is related to the arctangent. The logarithm is also multiple-valued when you do this kind of thing, since $\operatorname{Log} 1$ can be $0$ or $\pm2\pi i$ or $\pm4\pi i$, etc. ${}\qquad{}$ Commented Nov 7, 2015 at 20:20

Hint. Using partial fraction decomposition, you may prove that $$\frac{1}{1-2x^2}=\frac{1}{2 \left(1+\sqrt{2} x\right)}-\frac{1}{2 \left(-1+\sqrt{2} x\right)}$$
Hint : Write the integrand in the form $\frac{A}{1-\sqrt{2}z}+\frac{B}{1+\sqrt{2}z}$
Hint: if you need to use the arctan function in your integral, the solution is $$\int \frac{1}{1-2z^2}=\int \frac{1}{1+(\sqrt{2}iz)^2}dz=\frac{1}{\sqrt{2}i}\tan^{-1}\sqrt{2}iz+C$$
$$\int \frac{1}{1-2z^2} = \frac{1}{2} \int \frac{1}{1-\sqrt{2}z} + \frac{1}{1+\sqrt{2}z} = \frac{1}{2} (-\frac{1}{\sqrt{2}} \log|1-\sqrt{2}z| + \frac{1}{\sqrt{2}} \log|1+\sqrt{2}z|) = \frac{1}{2\sqrt{2}} \log|\frac{1+\sqrt{2}z}{1-\sqrt{2}z}|$$