$\int \frac{1}{{1-2x-x^2}} \, \mathrm{d}x $ substitution I have this integral.
$$\int \frac{1}{{1-2x-x^2}} \, \mathrm{d}x $$
But I am unable to do it right and I just don't know where is the problem in my steps.
My steps:  
Complete the square
$$\int \frac{1}{{2-(x+1)^2}} \, \mathrm{d}x $$
$$\frac{1}{2}*\int \frac{1}{{-(x+1)^2}} \, \mathrm{d}x $$
Substitute $$t=x+1$$
$$\frac{1}{2}*\int \frac{1}{{-t^2}} \, \mathrm{d}x $$
$$\frac{1}{2}*\int {{-t^{-2}}} \, \mathrm{d}x $$
$$\frac{1}{2}* \frac{-t^{{-1}}}{{-1}}+c  $$
$$ \frac{1}{{2(x+1)}}+c $$
But the result is different.
I will be thankful for any help.
 A: Hint:
$$1-2x-x^2$$ factors as
$$-(x-a)(x-b)$$ where
$$a,b=1\pm\sqrt2.$$
Then you can decompose
$$\frac{b-a}{(x-a)(x-b)}=\frac1{x-b}-\frac1{x-a}$$ and integrate
$$\log|x-b|-\log|x-a|.$$
A: Hint:
$$\int\dfrac{1}{1-2x-x^2}\mathrm dx=\dfrac{-1}{\sqrt{2}}\int\dfrac{1}{\left(\frac{x+1}{\sqrt{2}}\right)^2-1}\mathrm dx$$
This is inviting for Partial Fraction Decomposition. Can you proceed?

$$\int\dfrac{1}{2-(x+1)^2}\mathrm dx\ne\dfrac{1}{2}\int\dfrac{1}{-(x+1)^2}\mathrm dx$$
A: Use the u-substation $u=\frac{x+1}{\sqrt{2}}$ and then do a partial fraction decomposition on the fraction that you will get:
$$
\int\frac{1}{1-2x-x^2}\,dx=\int\frac{1}{2-(x+1)^2}\,dx=\\
\frac{\sqrt{2}}{2}\int\frac{1}{1-\left(\frac{x+1}{\sqrt{2}}\right)^2}\frac{d}{dx}\left(\frac{x+1}{\sqrt{2}}\right)\,dx=
\frac{\sqrt{2}}{2}\int\frac{1}{1-u^2}\,du=\\
\frac{\sqrt{2}}{2}\int\frac{1}{(1-u)(1+u)}\,du=
\frac{\sqrt{2}}{2}\int\left(\frac{1}{2}\frac{1}{1-u}-\frac{1}{2}\frac{1}{1+u}\right)\,du=\\
\frac{\sqrt{2}}{4}\left(-\ln{|1-u|-\ln{|1+u|}}\right)+C=\\
\frac{\sqrt{2}}{4}\ln{\left|\frac{1+u}{1-u}\right|}+C.
$$
And then do a back-substitution.
