Integrating $\int \frac{3x^4+2x^2+1}{\sqrt{x^4+x^2+1}} \mathrm{d}x$

Question

Evaluate the indefinite integral: $$\int \frac{3x^4+2x^2+1}{\sqrt{x^4+x^2+1}} \mathrm{d}x$$

I multiplied up and down by $x$ and substituted $x^6+x^4+x^2=t$ so that $2x(3x^4+2x^2+1)\mathrm{d}x=\mathrm{d}t$ and therefore the integral converts to $\displaystyle \int \frac{1}{2\sqrt{t}}\mathrm{d}t=\sqrt{t}+C$.

But the above approach was motivated by seeing the answer first. I couldn't solve it before that. What should be the more natural approach?

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Edit: To some, the above method may seem natural.Then can they provide an alternate method?

Answer 1

I want the power of denominator to be greater than numerator's by $1$(For $\dfrac{f'(x)}{f(x)}$ form). So I would multiply and divide $x^2$ inside the square root, to get the job done.

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Once you get it, multiply and divide some number $t$ to get it to its derivative form and balance it out by subtracting.$$\frac{1}{t}f'(x)-kg(x)=N^r$$

Coefficients are given with care in questions to avoid non integrable $g(x)$

Answer 2

You can use the following theorem by Charles Hermite.

If $P(t)$ is polynomial of degree $\geq 1$, then we can find $K, Q(t)$

$$\int\frac{P(t)}{\sqrt{at^2+bt+c}}dt = Q(t)\sqrt{at^2+bt+c} + \int \frac{K}{\sqrt{at^2+bt+c}}dt$$

your problem is convertible to this form by substitution $t=x^2, dt = 2x dx$

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J.G comments that we don't retain a polynomial after the substitution

But if we rewrite $$\int \frac 1 {2x} \cdot \frac{(3x^4+2x^2+1)2x}{\sqrt{x^4+x^2+1}}dx$$ and then do integration by parts, differentiating factor $\frac 1{2x}$. Not so sure it becomes easier, but you can use it.

Answer 3

Here's how I would have approached the problem, with all of the messiness included. We aim to evaluate $$\int \frac{3x^4+2x^2+1}{\sqrt{x^4+x^2+1}} \mathrm{d}x\text{.}$$

I don't like the $x^4$, $x^2$ in here, so let's try $u = x^2$. Then we have $$\int \dfrac{3u^2 + 2u + 1}{\sqrt{u^2 + u + 1}}\dfrac{\text{d}u}{2x}$$

There is a problem here: we have that both $x$ and $u$ are in the integral. So, we must attempt substitutions until we get this to work out. Given the $x$ is in the denominator, we probably want the numerator of the left fraction and the denominator of the right fraction to cancel out. So we must choose a suitable substitution for this to work.

To cancel out the left numerator, I need to choose a substitution $u(x)$ so that $$\dfrac{\text{d}u}{3x^4 + 2x^2 + 1} = \text{d}x$$ or $$\dfrac{\text{d}u}{\text{d}x} = 3x^4+2x^2+1\text{.}$$ This is a straightforward integration problem: we obtain $u(x) = \dfrac{3}{5}x^5+\dfrac{2}{3}x^3+x+ C$, and to simplify matters, I'll assume $C = 0$ for now. With our substitution, we obtain $$\int \dfrac{1}{\sqrt{x^4 + x^2 + 1}}\text{ d}u$$ and now I've created another problem for myself: I'm not sure what to do with the denominator of the left-hand side.

Two things are clear to me with this integral:

• The constants multiplying the $x^5$, $x^3$, etc. may work to simplify the numerator of the left fraction, but not the denominator.
• The powers that we've chosen in $u(x)$ above probably won't work, given that we're dealing with a square root. It suggests to me that we should be dealing with even powers, as opposed to odd powers.

Thus, in my opinion and given the above, it isn't too much of a stretch to think that $u(x) = x^6 + x^4 + x^2$ would be a suitable substitution for this integral.