How to integrate $\int \frac{4+\sqrt{x^{2}}}{2x^{4}+3x^{2}+1}dx$?

Question

How to integrate $$\int \frac{4+\sqrt{x^{2}}}{2x^{4}+3x^{2}+1}dx$$ ? I am trying to evaluate this integral by factorizing $2x^{4}+3x^{2}+1$. Now $2x^{4}+3x^{2}+1=(2x^{2}+1)(x^{2}+1)$. Therefore the above integral will become $$\int \frac{4+\sqrt{x^{2}}}{(2x^{2}+1)(x^{2}+1)}$$. Now I am integrating this expression by writing $$\int \frac{4}{(2x^{2}+1)(x^{2}+1)}dx$$ $$+$$ $$\int \frac{\sqrt{x^{2}}}{(2x^{2}+1)(x^{2}+1)}dx.$$ Now I am facing problem in integrating $$\int \frac{\sqrt{x^{2}}}{(2x^{2}+1)(x^{2}+1)}dx.$$ Because $\sqrt{x^{2}}=|x|$. Now $|x|=x(x>0),-x(x<0),0(x=0)$. Therefore we can write $|x|=xsgn(x)$. Therefore we can write the above integral as $$\int \frac{4}{(2x^{2}+1)(x^{2}+1)}dx$$ $$+$$ $$\int\frac{xsgn(x)}{(2x^{2}+1)(x^{2}+1)}dx.$$ But I can't proceed further. Please help me out with this integral.

Answer 1

Let $$J(x)=\int \frac{\sqrt{x^{2}}}{2x^{4}+3x^{2}+1}dx=\int \frac{|x|}{2x^{4}+3x^{2}+1}dx$$ If $x>0, J(x)=+K(x)$, if $x<0, J(x)=-K(x)$, where $$K(x)=\int \frac{x }{2x^{4}+3x^{2}+1}dx=\int \frac{x}{(2x^2+1)(x^2+1)}=\int \frac{xdx}{(2x^2+1)(x^2+1)}$$$$=\frac{1}{2}\int \left(\frac{dt}{2t+1}-\frac{dt}{t+1}\right), x^2=t.$$ Now you can take from here.

Answer 2

When $x\ge 0$ you have $$\frac{4+x}{2x^4+3x^2+1}=\frac{8+2x}{1+2x^2}-\frac{x+4}{1+x^2}.$$ For $x<0$ you instead get $$\frac{4-x}{2x^4+3x^2+1}=\frac{8-2x}{1+2x^2}+\frac{x-4}{1+x^2}.$$ You can find these alternate expressions using partial fraction decomposition since $$2x^4+3x^2+1=(1+x^2)(1+2x^2).$$

Now we know how to integrate anything of the form $$\frac{ax+b}{cx^2+dx+e}\quad(c\neq 0)$$ but takes a bit of work. This amounts to using combinations of $\ln$, $\arctan$, the power rule, and partial fractions. If the denominator is a linear factor squared, you can apply the power rule. If the denominator has distinct linear factors, use partial fractions. In this case all denominators are irreducible, so you will only need to use $\ln$ and $\arctan$. The idea is to use $\ln$ to take care of the $x$ term in the numerator, and then worry about the constant term. For example, consider $$\int\frac{8+2x}{1+2x^2}dx.$$ The derivative of $\ln|1+2x^2|$ is $\frac{4x}{1+2x^2}$, so $$\int\frac{2x}{1+2x^2}dx=\frac12\ln|1+2x^2|.$$ To find $\int\frac{8}{1+2x^2}dx$ you will have to use $\arctan$. Once you get the constants right, you find $$\int\frac{8}{1+2x^2}dx=4\sqrt 2\arctan(\sqrt 2x).$$ A similar process gives $$\int\frac{x+4}{1+x^2}dx=\frac12\ln|1+x^2|+4\arctan x.$$

Similarly $$\int\frac{x-4}{1+x^2}dx=\frac12\ln|1+x^2|-4\arctan x$$ and $$\int\frac{8-2x}{1+2x^2}dx=-\frac12\ln|1+2x^2|-4\sqrt 2\arctan(\sqrt 2x).$$