# Integration: $\;\int \frac{1}{2-x^2}\,dx$

Sorry to ask about a (probably) faily easy task, but I am new to integrals and can't solve this one:

I am trying to calculate $\displaystyle \int \frac{dx}{2-x^2}$.

The solution is $\displaystyle \frac{\log(x+\sqrt{2})-\log(\sqrt{2}-x)}{2\sqrt{2}}$, but I don't know how to get there.

It must have something to do with the rule $\displaystyle \int \frac{1}{x} = log(|x|)+c$.

I thought about writing $\displaystyle \int \frac{1}{(2-x^2)}$ as $\displaystyle \int (2-x^2)^{-1}$, but the rule for potences is not applicable here as it seems, because we would divide by zero.

So there must be multiple rules to get to the above term. Which ones would you use?

• Your answer isn't right, but close. Use partial fractions: $$\frac{1}{2-x^2} = \frac{A}{\sqrt{2}-x} + \frac{B}{\sqrt{2}+x}$$ – Simon S Jan 17 '15 at 16:54
• Than answer should have a $+$ or $-$ in the numerator, not the product of the logarithms. – Thomas Andrews Jan 17 '15 at 16:56

## 4 Answers

You can do partial fraction decomposition: $$\frac{ 1}{(2 - x^2)} =\frac A{\sqrt 2 - x} + \frac B{\sqrt 2 + x}$$

where $A(\sqrt 2+x) + B(\sqrt 2 - x) = 0\cdot x + 1$.

Then solve for $A, B$.

• $A = B = \frac{1}{2\sqrt{2}}$. Thanks that should work – Falco Winkler Jan 17 '15 at 17:04
• You're welcome, @Falco! – Namaste Jan 17 '15 at 17:05

HINT : Use partial Fraction and use resut you stated

• The more details you supply, the higher the chance you get upvotes. – DeepSea Jan 17 '15 at 16:57

$$\int \frac{1}{2-x^2}dx = \int \frac{1}{2} . \frac{1}{1-\frac{x^2}{2}}dx = \frac{1}{2}\int\frac{1}{1-\frac{x^2}{2}}dx$$

Now substitue $u=\frac{x}{\sqrt{2}}$ and $du=\frac{1}{\sqrt{2}}dx$.

Remember that

$$\int \frac{1}{a^2-x^2}dx=\frac{1}{2a} \ln \frac{|a+x|}{|a-x|}+C$$ for $x\neq a$

Hint: $\dfrac{1}{2-x^2} = \dfrac{1}{(\sqrt{2})^2 - x^2}= \dfrac{1}{2\sqrt{2}}\cdot \left(\dfrac{1}{\sqrt{2}-x}+\dfrac{1}{\sqrt{2}+x}\right)$