# How do you integrate int(dx/(sqrt(2x^3 + 2x + 5)))?

Apr 3, 2015

The integral you wrote is impossible to do using "elementary functions" (at least according to the computer program Mathematica...it requires "Elliptic Functions").

I think you probably meant to do $\setminus \int \left(\setminus \frac{\mathrm{dx}}{\setminus \sqrt{2 {x}^{2} + 2 x + 5}}\right)$. The first steps in doing this integral is to use a bit of tricky algebra (including the technique of completing the square) to get:

$\setminus \int \setminus \frac{\mathrm{dx}}{\setminus \sqrt{2 {x}^{2} + 2 x + 5}} = \setminus \frac{1}{\setminus \sqrt{2}} \setminus \int \setminus \frac{\mathrm{dx}}{\setminus \sqrt{{x}^{2} + x + \setminus \frac{5}{2}}}$

$= \setminus \frac{1}{\setminus \sqrt{2}} \setminus \int \setminus \frac{\mathrm{dx}}{\setminus \sqrt{{x}^{2} + x + \setminus \frac{1}{4} + \setminus \frac{9}{4}}} = \setminus \frac{1}{\setminus \sqrt{2}} \setminus \int \setminus \frac{\mathrm{dx}}{\setminus \sqrt{{\left(x + \setminus \frac{1}{2}\right)}^{2} + \setminus \frac{9}{4}}}$.

Now do the trigonometric substitution $x + \setminus \frac{1}{2} = \setminus \frac{3}{2} \setminus \tan \left(\setminus \theta\right)$ do so that $\setminus \theta = \setminus \arctan \left(\setminus \frac{2}{3} x + \setminus \frac{1}{3}\right)$ and $\mathrm{dx} = \setminus \frac{3}{2} \setminus {\sec}^{2} \left(\setminus \theta\right) d \setminus \theta$.

Then the integral becomes:

$\setminus \frac{1}{\setminus \sqrt{2}} \setminus \int \setminus \frac{\mathrm{dx}}{\setminus \sqrt{{\left(x + \setminus \frac{1}{2}\right)}^{2} + \setminus \frac{9}{4}}} = \setminus \frac{1}{\setminus \sqrt{2}} \setminus \int \setminus \frac{\setminus \frac{3}{2} \setminus {\sec}^{2} \left(\setminus \theta\right) d \setminus \theta}{\setminus \sqrt{\setminus \frac{9}{4} \left(\setminus {\tan}^{2} \left(\setminus \theta\right) + 1\right)}}$, which simplifies to

$\setminus \frac{1}{\setminus \sqrt{2}} \setminus \int \setminus \frac{\setminus {\sec}^{2} \left(\setminus \theta\right) d \setminus \theta}{\setminus \sqrt{\setminus {\sec}^{2} \left(\setminus \theta\right)}}$

after using a trigonometric identity. Technically-speaking, $\setminus \sqrt{\setminus {\sec}^{2} \left(\setminus \theta\right)} = | \sec \left(\setminus \theta\right) |$, however, it is fine to get rid of the absolute value signs for the purpose of finishing the integral (assume $\setminus \sec \left(\setminus \theta\right) \setminus \ge q 0$).

Hence, the integral becomes $\setminus \frac{1}{\setminus \sqrt{2}} \setminus \int \setminus \sec \left(\setminus \theta\right) d \setminus \theta$. This last integral looks simple, but it is still tricky to do. One trick that works is to multiply it by \frac{\sec(\theta)+\tan(\theta}}{\sec(\theta)+\tan(\theta) and then do a substitution (you can try that if you want). Wolfram Alpha gives the answer as $\setminus \frac{1}{\setminus \sqrt{2}} \left(\ln \left(\cos \left(\setminus \frac{\theta}{2}\right) + \sin \left(\setminus \frac{\theta}{2}\right)\right) - \setminus \ln \left(\setminus \cos \left(\setminus \frac{\theta}{2}\right) - \setminus \sin \left(\setminus \frac{\theta}{2}\right)\right)\right) + C$. Now just replace $\setminus \theta$ by $\setminus \theta = \setminus \arctan \left(\setminus \frac{2}{3} x + \setminus \frac{1}{3}\right)$ and you are done:

$\setminus \int \left(\setminus \frac{\mathrm{dx}}{\setminus \sqrt{2 {x}^{2} + 2 x + 5}}\right) = \setminus \frac{1}{\setminus \sqrt{2}} \left(\ln \left(\cos \left(\setminus \frac{1}{2} \setminus \arctan \left(\setminus \frac{2}{3} x + \setminus \frac{1}{3}\right)\right) + \sin \left(\setminus \frac{1}{2} \setminus \arctan \left(\setminus \frac{2}{3} x + \setminus \frac{1}{3}\right)\right)\right) - \setminus \ln \left(\setminus \cos \left(\setminus \frac{1}{2} \setminus \arctan \left(\setminus \frac{2}{3} x + \setminus \frac{1}{3}\right)\right) - \setminus \sin \left(\setminus \frac{1}{2} \setminus \arctan \left(\setminus \frac{2}{3} x + \setminus \frac{1}{3}\right)\right)\right)\right) + C$

It is possible to simplify this by using half angle formulas and the following equations, but it may not be worth the trouble: $\setminus \cos \left(\setminus \arctan \left(\setminus \frac{2}{3} x + \setminus \frac{1}{3}\right)\right) = \setminus \frac{1}{\setminus \sqrt{1 + {\left(\setminus \frac{2}{3} x + \setminus \frac{1}{3}\right)}^{2}}} = \setminus \frac{3}{\setminus \sqrt{4 {x}^{2} + 4 x + 10}}$

$\setminus \sin \left(\setminus \arctan \left(\setminus \frac{2}{3} x + \setminus \frac{1}{3}\right)\right) = \setminus \frac{\setminus \frac{2}{3} x + \setminus \frac{1}{3}}{\setminus \sqrt{1 + {\left(\setminus \frac{2}{3} x + \setminus \frac{1}{3}\right)}^{2}}} = \setminus \frac{2 x + 1}{\setminus \sqrt{4 {x}^{2} + 4 x + 10}}$