# Where was my mistake (integration by trig-substitution problem)?

I am attempting to solve the problem

$$\int \frac{dx}{x^2+x+1}$$

First, I complete the square, then factor out a $\frac {3}{4}$:

$$\int \frac{dx}{\frac{3}{4}(\frac{4}{3}(x+\frac{1}{2})^2+1)}$$

Let $u = \sqrt{\frac{4}{3}}(x+\frac{1}{2})$

$$\frac{du}{dx} = \frac{2}{\sqrt{3}}$$

$$dx = \frac{\sqrt{3}}{2} du$$

Thus, we now have the integral:

$$\frac{4}{3} \frac{\sqrt{3}}{2} \int \frac{du}{u^2+1}$$

Let $u = \tan \theta$

$$du = \sec^2\theta \ d\theta$$

What follows is obvious now, and the solution should be:

$$\frac{4}{3} \frac{\sqrt{3}}{2} \theta + C$$

$$\theta = \tan^{-1}(u)$$

Thus, the final solution is:

$$\frac{4}{3} \frac{\sqrt{3}}{2} \tan^{-1} \left( \sqrt{\frac 4 3} \left(x+\frac 1 2 \right)\right) + C$$

However, according to online calculator integral-calculator, the answer is:

$$\frac 2 {\sqrt 3} \tan^{-1} \left( \frac{2x+1}{\sqrt 3} \right)+C$$

Any indication as to where my mistake falls would be very beneficial.

• The two answers you've given are equal. You just need to do some algebra to see it. For instance, $\frac{4}{3} \frac{\sqrt{3}}{2} = \frac{2}{\sqrt{3}}$ and $\frac{2}{\sqrt{3}}(x + 1/2) = \frac{2x+1}{\sqrt{3}}$. Jul 5, 2017 at 4:48
• @Quasicoherent Ah, I see. The calculator said they weren't but that its answer checking service is unreliable. Thank you! Jul 5, 2017 at 4:49
• Possible duplicate of Integral of $\frac{1}{x^2+x+1}$ and$\frac{1}{x^2-x+1}$
– zwim
Jul 5, 2017 at 5:26
• @zwim Most certainly not a duplicate. Here the OP has already figure out how to find the integral, and is only having problems simplifying his answer to match what we was supposed to get
– Ant
Jul 5, 2017 at 8:10

$$x^2+x+1=(x+1/2)^2+3/4\ne \frac34 \left(\frac43 (x+1/2)^2+\frac34 \right)=x^2+2+5/2$$
• My apologies, I meant for it to be $\int \frac{dx}{\frac{3}{4}(\frac{4}{3}(x+\frac{1}{2})^2+1)}$ Jul 5, 2017 at 4:49
Your only mistake appears to be a failure to notice that $\displaystyle \frac{4}{3} \frac{\sqrt{3}}{2} \tan^{-1} \left( \sqrt{\frac 4 3} \left(x+\frac 1 2 \right)\right)$ is exactly the same thing as $\displaystyle \frac 2 {\sqrt 3} \tan^{-1} \left( \frac{2x+1}{\sqrt 3} \right).$
First you have $$\frac 4 3 \cdot \frac{\sqrt 3} 2 = \frac{4\sqrt 3}{\sqrt 3\sqrt 3 \cdot 2} = \frac 2 {\sqrt 3}.$$ And then $$\sqrt{\frac 4 3} \left( x + \frac 1 2 \right) = \frac 2 {\sqrt 3} \left( x + \frac 1 2 \right) = \frac 1 {\sqrt 3} (2x+1).$$