Solving the the Integral: $\int\frac{1}{u^2-2}du$ I am attempting to solve the following integral, in which I am not sure, how to proceed:
$\int\frac{1}{u^2-2}du$
I have tried the following, but I am not sure if it is correct:
$y = u^2-2$
$dy = 2u*du$
$du = \frac{dy}{2u}$
Therefore:
$\int\frac{1}{u^2-1}du = \int\frac{1}{y}*\frac{dy}{2u} = \int\frac{1}{y}*\frac{1}{2u}dy = \int\frac{1}{y2u}dy$
Solving for u:
$u^2 = y+2$
$u = \sqrt{y + 2}$
Substituting $u$ for $\sqrt{y + 2}$:
$\int\frac{1}{y2u}dy = \int\frac{1}{y2(\sqrt{y + 2})}dy$
I am not sure, where to go from here. The only thing I can remember is to use the logarithm rule where:
$\int\frac{1}{y2(\sqrt{y + 2})}dy = ln|{y2(\sqrt{y + 2})}| + C$
Would that be correct?
 A: $$\int\frac{1}{u^2-2}du=\frac{1}{2\sqrt2}\int\left(\frac{1}{u-\sqrt2}-\frac{1}{u+\sqrt2}\right)du=\frac{1}{2\sqrt2}\ln\left|\frac{u-\sqrt2}{u+\sqrt2}\right|+C$$
A: hint
Begin by the substitution $$u=\sqrt {2}v $$
the integral becomes
$$\int \frac {\sqrt {2}dv}{2 (v^2-1)}=\frac {1}{\sqrt {2}}\int\frac {dv}{v^2-1} $$
then use
$$\frac {1}{v^2-1}=\frac {1}{2}\Bigl (\frac {1 }{v-1}-\frac {1}{v+1}\Bigr) $$
A: HINT:
$$
\frac{1}{u^2-2}=\frac{1}{(u-\sqrt2)(u+\sqrt2)}
$$
If you proceed with the scomposition for your algebraic fraction, you have:
$$
\frac{1}{(u-\sqrt2)(u+\sqrt2)}=\frac{A}{u-\sqrt2}+\frac{B}{u+\sqrt2}=\frac{A(u+\sqrt2)+B(u-\sqrt2)}{(u-\sqrt2)(u+\sqrt2)}
$$
You have that $A+B=0 \ $  AND $\ A-B=\sqrt2.\ $ So $A=\frac{\sqrt2}{2}$ and $B=- \frac{\sqrt2}{2}.$
Doing this, we come to the following integrals:
$$
\frac{\sqrt2}{2}\int\frac{\sqrt2}{u-\sqrt2}du - \frac{\sqrt2}{2}\int\frac{\sqrt2}{u+\sqrt2}du
$$
So, let's say the solutions..
A: Another approach: Let $u=\sqrt2\sec t$, so $du=\sqrt2\sec t\tan t\,dt$. Then we have:
$$\begin{align}
\int\frac1{u^2-2}du &= \int\frac{\sqrt2\sec t\tan t}{2\sec^2t-2}dt\\
&= \frac1{\sqrt2}\int\frac{\sec t \tan t}{\sec^2 t-1}dt\\
&= \frac1{\sqrt2}\int\frac{\sec t\tan t}{\tan^2 t}dt\\
&= \frac1{\sqrt2}\int\frac{\sec t}{\tan t}dt\\
&= \frac1{\sqrt2}\int\csc t\,dt\\
&= -\frac1{\sqrt2}\ln\left|\csc t + \cot t\right|+C
\end{align}$$
Since $\sec t=\frac{u}{\sqrt2}$, we have $\csc t=\frac{u}{\sqrt{u^2-2}}$ and $\cot t=\frac{\sqrt2}{\sqrt{u^2-2}}$. Thus, we end up with:
$$\begin{align}
-\frac{1}{\sqrt2}\ln\left|\frac{u+\sqrt2}{\sqrt{u^2-2}}\right|+C &= -\frac{1}{\sqrt2}\ln\left|\sqrt{\frac{(u+\sqrt2)^2}{u^2-2}}\right|+C\\
&= -\frac{1}{2\sqrt2}\ln\left|\frac{(u+\sqrt2)^2}{(u+\sqrt2)(u-\sqrt2)}\right|+C\\
&= -\frac{1}{2\sqrt2}\ln\left|\frac{u+\sqrt2}{u-\sqrt2}\right|+C\\
&= \frac{1}{2\sqrt2}\ln\left|\frac{u-\sqrt2}{u+\sqrt2}\right|+C
\end{align}$$
Messy, but possibly instructive. Partial fractions are cleaner than trig substitution, when you can use them.
A: An important method is the decomposition in simple fractions, which you can use when the integrand is the quotient of two polynomials and you can factor the denominator.
Here $u^2-2=(u-\sqrt2)(u+\sqrt2)$ and the decomposition is of the form
$$\frac a{u-\sqrt2}+\frac b{u+\sqrt2}=\frac{au+a\sqrt2+bu-\sqrt2b}{u^2-2}.$$
Then with $a=-b=\dfrac1{\sqrt8}$ we have the identity 
$$\frac1{\sqrt8(u-\sqrt2)}-\frac1{\sqrt8(u+\sqrt2)}=\frac1{u^2-2}.$$
Then you finish with the well-known
$$\int\frac{du}{u+a}=\ln|u+a|.$$
A: $$\int\frac{du}{u^2-2}$$
Using trig substitution we have
$$u=\sqrt2 {\sec x}$$
$$du=\sqrt2 \sec x \tan x dx$$
$$\int \frac{\sqrt2 \sec x \tan x dx}{2\tan^2x}=\int \frac{1}{\sqrt2}\frac{dx}{\sin x}$$
$$\int \frac{1}{\sqrt2}\frac{dx}{\sin x}=\frac{1}{\sqrt 2}\int \csc x dx$$
$$-\frac{1}{\sqrt 2} \ln|\csc x + \cot x| + C$$
Now just substitute.
