I'm working on a calculus problem and need help solving the following definite integral:

integral in question

I'm struggling to simplify the integrand or find a substitution that makes the integral easier to evaluate. Here's what I've tried so far:

  • Attempt at simplifying the inner square root.
  • Checking if the function is even to potentially simplify the limits of integration.

Unfortunately, I haven't made much progress. Does anyone have suggestions on techniques or substitutions that might work for this integral? Any guidance or references to similar problems would be greatly appreciated. Thanks.

the definite integral:

$$ \int_{-1/2}^{1/2} \sqrt{x^2 + 1 + \sqrt{x^4 + x^2 + 1}} \, dx $$

  • 1
    $\begingroup$ I'm not sure that the indefinite integral of this function exists in terms of standard functions $\endgroup$
    – ράτ
    Apr 21 at 11:43
  • $\begingroup$ @AntonyTheo Wolfram alpha would agree with you. it estimates the value of it at $\approx 1.45...$, but fails to find an antiderivative. $\endgroup$
    – B.A.M
    Apr 21 at 12:13
  • $\begingroup$ The solution is not that complicated though , I feel like wolfram alpha should be able to find it, maybe not have a step-by-step solution too but at least find the antiderivative. $\endgroup$
    – ράτ
    Apr 21 at 12:47

1 Answer 1


First notice that $x^4+x^2+1=(x^2+1-x)(x^2+1+x)$


$$I= \int_{-1/2}^{1/2} \sqrt{x^2 + 1 + \sqrt{(x^2+1-x)(x^2+1+x)}} \ dx$$

We can multiply and devide by $2$

$$I= \int_{-1/2}^{1/2} \sqrt\frac{2({x^2 + 1 + \sqrt{(x^2+1-x)(x^2+1+x)})}} 2 \ dx \\\int_{-1/2}^{1/2} \sqrt\frac{2{x^2 + 2 +2 \sqrt{(x^2+1-x)(x^2+1+x)}}} 2 \ dx =\int_{-1/2}^{1/2}\sqrt{\frac{1}{2}}\sqrt{2x^2+2+2\sqrt{(x^2+x+1)(x^2-x+1)}} \ dx $$

Then we can see that $\sqrt{2x^2+2+2\sqrt{(x^2+x+1)(x^-x+1)}} $ can be simplified to $\sqrt{x^2+x+1}+\sqrt{x^2-x+1}$

Replacing the expression with the simplified form we get

$$\int_{-1/2}^{1/2}\sqrt{\frac{1}{2}}(\sqrt{x^2+x+1}+\sqrt{x^2-x+1})dx \\ =\sqrt{\frac{1}{2}}\int_{-1/2}^{1/2}(\sqrt{x^2+x+1}+\sqrt{x^2-x+1})dx$$

Then we can see that $\sqrt{x^2+x+1}+\sqrt{x^2-x+1}$ is an even function so we rewrite the integral as

$$I=2\sqrt{\frac{1}{2}}\int_{0}^{1/2}(\sqrt{x^2+x+1}+\sqrt{x^2-x+1})dx =\\2\sqrt{\frac{1}{2}}(\int_{0}^{1/2}(\sqrt{x^2+x+1})dx + \int_{0}^{1/2}(\sqrt{x^2-x+1})dx $$

Let $I_1=\int_{0}^{1/2}(\sqrt{x^2+x+1})dx$ and $I_2=\int_{0}^{1/2}(\sqrt{x^2-x+1})dx$

Concider $I_1$


To evaluate this integral we let $u=x+\frac{1}{2}$

$$I_1= \int_{1/2}^{1}(\sqrt{u^2+\frac{3}{4}})du$$

Then substitute $u=\frac{\sqrt{3}}{2} \tan(s) $

$$I_1=\int ^{\tan^{-1}(2/ \sqrt{3})}_{π/6} \sqrt{(\frac{\sqrt{3}}{2} \tan(s))^2+ \frac{3}{4}\sec^2(s)}ds$$

Simplify the inside and we get

$$I_1=\int ^{\tan^{-1}(2/ \sqrt{3})}_{π/6}\frac{3\sec^3(s)}{4}ds\\=\frac{\sqrt{7}-1}{4}+ \frac{1}{8} \log(\frac{19\sqrt{7}+50}{27})\\ = \frac{\sqrt{7}}{4}+\frac{3}{8} \log(\frac{2+\sqrt{7}}{\sqrt{3}})-\frac{1}{4}-\frac{3}{16}\log(3) $$

Now consider $I_2$

we can see $x^2-x+1= (x- \frac{1}{2})^2+ \frac{3}{4}$

To evaluate the integral we substitute $u=x-\frac{1}{2}$

$$I_2=\int_0^{1/2} \sqrt{(x- \frac{1}{2})^2+ \frac{3}{4}}dx =\int^0_{-1/2} \sqrt{u^2+ \frac{3}{4}}du$$

Substituting $\frac{\sqrt{3}}{2}\tan(s)=u $

$$\int^0_{-π/6}\sqrt{(\frac{\sqrt{3}}{2}\tan(s))^2}+\frac{3}{4} \frac{\sqrt{3}}{2}\sec^2(s) ds\\ =\int^0_{-π/6}\frac{3\sec^3(s)}{4}ds= \frac{1}{4}+ \frac{3}{16}\log(3) $$

Then finally

$$I=2\sqrt{\frac{1}{2}}(I_1+I_2)\\ =2\sqrt{\frac{1}{2}}( \frac{\sqrt{7}}{4}+\frac{3}{8} \log(\frac{2+\sqrt{7}}{\sqrt{3}})-\frac{1}{4}-\frac{3}{16}\log(3) + \frac{1}{4}+ \frac{3}{16}\log(3) ) \\ =\frac{\sqrt{7}}{2 \sqrt{2}}+\frac{3}{4 \sqrt{2}}\log(\frac{2+\sqrt{7}}{\sqrt{3}}) $$

$$I=\frac{\sqrt{7}}{2\sqrt{2}}+\frac{3}{4\sqrt{2}}\log(\frac{2+\sqrt{7}}{\sqrt{3}}) \approx 1.45866$$

$\log$ is referring to the natural logarithm


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .