Help with the indefinite integral $\int \frac{dx}{2x^4 + 3x^2 + 5}$

Question

I start by rewriting the denominator, $2x^4+3x^2+5$, as a squared term plus a constant. To do this, we notice that the first two terms already have a common factor of $2x^2$. We can complete the square by taking half of the coefficient of our $x^2$ term, squaring it, and adding it to both the numerator and denominator.

Since the coefficient of our $x^2$ term is $3$, half of it would be $\frac{3}{2}$, and squaring it gives us $\frac{4}{9}$. So we add $\frac{4}{9}$ to both the numerator and denominator.

I started like that but I couldn't go further. Can you please help me to solve it.

Answer 1

Substitute $t=\sqrt[4]{\frac25}x$, along with $a_{\pm}= \sqrt{ 2\pm\frac3{\sqrt{10}}}$ \begin{align} &\int\frac1{2x^4+3x^2+5}dx = \frac1{\sqrt[4]{250}}\int\frac1{t^4+\frac3{\sqrt{10}} t^2 +1}dt\\ =&\ \frac1{2\sqrt[4]{250}}\int\frac{1+t^2}{t^4+\frac3{\sqrt{10}} t^2 +1} + \frac{1-t^2}{t^4+\frac3{\sqrt{10}} t^2 +1}\ dt\\ =&\ \frac1{2\sqrt[4]{250}}\int\frac{d(t-\frac1t)}{(t-\frac1t)^2+a_+^2} -\frac{d(t+\frac1t)}{(t+\frac1t)^2-a_-^2}\\ =&\ \frac1{2\sqrt[4]{250}} \bigg( \frac1{a_+}\tan^{-1}\frac{t-\frac1t}{a_+} + \frac1{a_-}\coth^{-1}\frac{t+\frac1t}{a_-}\bigg) \end{align}

Answer 2

Another solution

Write $$2 x^4+3 x^2+5=2(x^2-a)(x^2-b)$$ with $$a=-\frac{1}{4} \left(3+i \sqrt{31}\right) \qquad \text{and} \qquad b=-\frac{1}{4} \left(3-i \sqrt{31}\right) $$ Use partial fraction decomposition $$\frac 1 {2 x^4+3 x^2+5}=\frac{1}{2 (a-b)}\left(\frac{1}{x^2-a}-\frac{1}{x^2-b} \right)$$ So, two simple integrals