Integration of a rational function involving a quadratic 
Evaluate the integral: $$\int\frac{3x}{(1-4x-2x^2)^2}\ dx$$

Here is my work:

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*Complete the square on the denominator: $$(1-4x-2x^2)=(1-2(2x+x^2+1-1))=(3-2(x+1)^2)$$


*Insert back into denominator. Use substitution $u=x+1$; $x=u-1$; $dx=du$. $$\int\frac{3(u-1)}{(3-2u^2)^2}\ du$$


*Split the integral into two: $$\int\frac{3u}{(3-2u^2)^2}\ du\ -\int\frac3{(3-2u^2)^2}\ du$$


*Integrate the first integral: $$\int\frac{3u}{(3-2u^2)^2}\ du$$
$v=(3-2u^2)$
$\frac{dv}{du}=-4u$
$\frac{-dv}4=u\ du$
$$\frac{-1}4\int\frac{3}{v^2}\ dv=\frac{-3}{4}\int\frac{1}{v^2}\ dv=\frac1{4v}$$$$=\frac{1}{4(3-2u^2)}=\frac1{4(3-2(x+1)^2)}$$

*

*Integrate the second integral: $$-\int\frac3{(3-2u^2)^2}\ du$$
Use sine trig substitution: $u=\frac{\sqrt3}{\sqrt2}\sin\theta$; $\frac{\sqrt2}{\sqrt3}du=\cos\theta\ d\theta$
$$-\frac{\sqrt2}{\sqrt3}\int\frac{3\cos\theta}{(3-2(\frac{\sqrt3}{\sqrt2}\sin\theta)^2)^2}\ d\theta=-\frac{\sqrt2}{\sqrt3}\int\frac{3\cos\theta}{(3-2(\frac{3}{2}\sin^2\theta))^2}\ d\theta$$
$$=-\frac{\sqrt2}{\sqrt3}\int\frac{3\cos\theta}{(3-3\sin^2\theta))^2}\ d\theta=-\frac{\sqrt2}{3\sqrt3}\int\frac{\cos\theta}{(\cos^2\theta))^2}\ d\theta=-\frac{\sqrt2}{3\sqrt3}\int\frac{1}{(\cos^3\theta)}\ d\theta$$$$=-\frac{\sqrt2}{3\sqrt3}\int\sec^3\theta\ d\theta=\frac{\sqrt2}{3\sqrt3}\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|$$
Besides for the last resubstitution which I still have to do, I also seem to have a different leading fraction for my second integral than the answer key. Where did I go wrong?

Hint:
This formula should be known by heart:
$$\int\!\frac {\mathrm d x}{a^2-x^2}=\frac 1a\,\operatorname{argtanh}\left(\frac xa\right)=\frac 1{2a} \ln\left(\frac{a+x}{a-x}\right).$$
(Similar to $\;\displaystyle\int\!\frac {\mathrm d x}{a^2+x^2}=\frac 1a\,\arctan\left(\frac xa\right) )$
