# Integration of a rational function involving a quadratic

Evaluate the integral: $$\int\frac{3x}{(1-4x-2x^2)^2}\ dx$$

Here is my work:

• 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?

• you inverted the $\sqrt{2}/\sqrt{3}$ when substituting $u$ by $\theta$. – Daniel Sep 20 '19 at 0:23
• don't I need to do that though in order to get it on the side of the du? – Burt Sep 20 '19 at 0:25
• Notice you wrote $\frac{\sqrt2}{\sqrt3}du=\cos\theta\ d\theta$, and so it must be $du=\frac{\sqrt3}{\sqrt2}\cos\theta\ d\theta$. – Daniel Sep 20 '19 at 0:28

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 $$\;\displaystyle\int\!\frac {\mathrm d x}{a^2+x^2}=\frac 1a\,\arctan\left(\frac xa\right) )$$
Here is how to carry out the second integral. With the substitution $$u=\frac{\sqrt3}{\sqrt2}\sin\theta$$,
$$I=-\int\frac3{(3-2u^2)^2}\ du$$
$$=-\frac{\sqrt3}{\sqrt2}\int\frac{3\cos\theta}{(3-2(\frac{3}{2}\sin^2\theta))^2}\ d\theta$$ $$=-\frac{\sqrt3}{\sqrt2}\int\frac{3\cos\theta}{(3-3\sin^2\theta)^2}\ d\theta =-\frac{\sqrt3}{3\sqrt2}\int\frac{\cos\theta}{\cos^4\theta}\ d\theta$$ $$=-\frac{1}{\sqrt6}\int\frac{1}{\cos^3\theta}\ d\theta =-\frac{1}{\sqrt6}\int\sec^3\theta\ d\theta$$ $$=-\frac{1}{2\sqrt6}(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)+C$$