# Help evaluating $\int \sqrt{{x}^{2} + 3} \; dx$

Can you help me evaluating the following indefinite integral?

$$\int \sqrt{{x}^{2} + 3} \; dx$$

Please, don't give a full solution, just some hint on which method to use...

** UPDATE **

Thank you very much to everybody for the useful comments and suggestions. I'm sorry for the delay with my reply, unfortunately do some mathematics as an hobby and often don't have time to work at it.

I tried to take Lucian suggestion on board and use trigonometric substitution as follows.

$$x = \sqrt{3} \tan\theta$$

and

$$\int \sqrt{{x}^{2} + 3} \; dx = \int \sqrt{{3\tan}^{2}\theta + 3} \; \sqrt{3}\sec^{2}\theta \; d\theta = \int \sqrt{{3\sec}^{2}\theta} \; \sqrt{3}\sec^{2}\theta \; d\theta = \int \sqrt{3}\sec\theta \; \sqrt{3}\sec^{2}\theta \; d\theta = 3 \int \sec^{3}\theta \; d\theta$$

which (according to common integral tables) is equal to

$$3 \left[ \frac{1}{2} \sec\theta \tan\theta + \frac{1}{2} \ln\left| \sec\theta + \tan\theta \right| + C \right]$$

Problem arises when I try to substitute back the variable from $$\theta$$ to $$x$$ because I know $$\tan\theta = \frac{x}{\sqrt{3}}$$ but I don't know how to substitute back $$\sec\theta$$, so basically I stopped here:

$$\frac{3}{2}\sec\theta\;\frac{x}{\sqrt{3}} + \frac{3}{2}\ln\left| \sec\theta + \frac{x}{\sqrt{3}} \right| + 3C$$

It looks close to the final answer but still not there...any suggestions?

• Replace $x$ with $\sqrt{3} \sinh u$. Mar 29, 2015 at 15:18
• or Integrate by parts Mar 29, 2015 at 15:26
• @suhail, what do you after the first step?
– abel
Mar 29, 2015 at 15:29
• @abel See my answer. I hope that it is useful Mar 29, 2015 at 15:56
• @Jashin I really would like to help, so please let me know how I can improve my answer. I just want to give you the best answer I can. Mar 29, 2015 at 15:59

One may recall that $$\cosh^2 u -\sinh^2 u=1 \quad \text{or}\quad \cosh^2 u =1+\sinh^2 u,$$ then you may try the change of variable $$x=\sqrt{3}\sinh u, \quad dx=\sqrt{3}\cosh u\: du,$$ giving \begin{align} \int\sqrt{x^2+3}\:dx &=\int\sqrt{3\:(\sinh^2 u+1)}\times \sqrt{3}\cosh u\: du\\\\ &=3\int\cosh^2 u \:du\\\\ &=\frac{3}2\int\left(1+\cosh(2 u)\right) \:du. \end{align} Hoping you can take it from here.

Please, don't give a full solution, just some hint on which method to use...

Hint: Let $x^2=3\tan^2t$, and use the fact that $\tan't=1+\tan^2t=\dfrac1{\cos^2t}$

You have modified your question and now you want to determine the substitution for $$\sec\theta$$. So, here we go:

Given: $$\tan\theta = \frac{x}{\sqrt{3}}$$ knowing that $$\tan\theta$$ is positive for positive $$x$$, squaring both sides: $$tan^2\theta = \frac{x^2}{3}$$ $$1+ tan^2\theta = 1+ \frac{x^2}{3}$$ $$1+ \frac{sin^2\theta}{cos^2\theta} = 1+ \frac{x^2}{3}$$ $$\frac{cos^2\theta + sin^2\theta}{cos^2\theta} = 1+ \frac{x^2}{3}$$ and since $$cos^2\theta + sin^2\theta = 1$$, we have: $$\frac{1}{cos^2\theta} = 1+ \frac{x^2}{3}$$ $$\sec^2\theta = 1+ \frac{x^2}{3}$$ $$\sec\theta = \sqrt{1+ \frac{x^2}{3}}$$

This is your substitution. One would argue that how did I remove squares from both sides, isn't that illegal? Well, not here - since I was the one who squared the sides in the first place, I know that the positive square root is the correct one, as the negative one was the result of me squaring the equation and creating an additional root of the equation. The positive root is consistent with the equation I started with, hence it's the one we are looking for.

Just a hint. A possible approach would be to consider using the formula for the integral of an irrational function like this: $$r={\sqrt {a^{2}+x^{2}}}$$ so that $$\int r\;dx={\frac {1}{2}}\left(xr+a^{2}\,\ln \left({x+r}\right)\right) + C$$. For restricted x values it can be represented as an expression with $$sinh$$.

• Your Answer is also good. But maybe it would have been more useful if you could have given him what OP required. Nov 12, 2019 at 13:05

Substitution $$\sqrt{x^2+3}=u-x$$ will work

$$\sqrt{x^2+3}=u-x\\ x^2 +3 = u^2 - 2ux + x^2\\3=u^2-2ux \\ 2ux = u^2-3 \\ x = \frac{u^2-3}{2u} \\ u - x = u - \frac{u^2-3}{2u}\\u-x=\frac{u^2+3}{2u} \\ dx = \frac{2u\cdot2u-2(u^2-3)}{4u^2}du \\ dx = \frac{2u^2+6}{4u^2}du \\ dx = \frac{u^2+3}{2u^2}du \\ \int{\frac{u^2+3}{2u^2}\cdot\frac{u^2+3}{2u}du} \\ \frac{1}{4}\int{\frac{u^4+6u^2+9}{u^3}du}\\ \frac{1}{4}\left(\frac{1}{2}u^2-\frac{9}{2}\frac{1}{u^2}\right)+\frac{3}{2}\ln{|u|}+C\\\frac{1}{2}(\frac{u^4-9}{4u^2})+\frac{3}{2}\ln{|u|}+C\\ \frac{1}{2}\frac{u^2-3}{2u} \cdot \frac{u^2+3}{2u}+\frac{3}{2}\ln{|u|}+C \\ \frac{1}{2}x\sqrt{x^2+3} + \frac{3}{2}\ln{|x+\sqrt{x^2+3}|}+C$$

$$x=\sqrt{3}\tan{\theta}\\ \frac{x}{\sqrt{3}} = \tan{\theta}$$
Recall identity $$1+\tan^2{\theta}=\sec^2(\theta)$$
$$\frac{x^2}{3}=\tan^2{\theta}\\ 1+\frac{x^2}{3} = 1+ \tan^2{\theta} \\ 1+\frac{x^2}{3} = \sec^2{\theta}\\ \sqrt{\frac{3+x^2}{3}} = \sec{\theta}$$
but you must be sure that secant is nonnegative

In my opinion subsitution which i showed is better for this integral
because after this substitution we only need to integrate power function