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

Substitute $u= \frac{2x-1}{\sqrt{3}}, du=\frac{2}{\sqrt{3}}dx$:

$$\frac {\sqrt{3}} 2 \int \frac{\frac{\sqrt{3}} {2}u + \frac 1 2}{(\frac{\sqrt{3}}{2}u)^2+(\frac {\sqrt{3}}{2})^2} = \int \frac{u}{u^2+1}du + \frac{1}{\sqrt{3}}\int\frac 1 {u^2+1}du.$$

This gives $$\frac 1 2\log({u^2+1})+\frac{1}{\sqrt{3}}\arctan{u}.$$ Substituting back in x yields $$\frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3)+\frac 1 {\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})$$

However, according to Wolfram Alpha, the integral should evaluate to $$\frac 1 2 \log(x^2-x+1)+\frac{1}{\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})$$After working some time on the integral, I know how to reach this solution, but I don't understand why my first attempt didn't arrive at the correct answer. Do you see where I went wrong?

Thank you for any help!

Hans Engler
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beforepim
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3 Answers3

7

Hello and welcome to math.stackexchange. The solution that you developed and the one given by Wolfram Alpha only differ by a constant (of integration), since
$$ \frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3) = \frac 1 2\log(x^2- x+ 1) + \frac 1 2 \log \frac 4 3 \, . $$

Hans Engler
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5

Both of the answers are correct. You have forgotten to add the integration constants in the solutions.

The first solution $\dfrac 12 \ln \left(\dfrac 43 x^2 -\dfrac 43 x +\dfrac 43\right) +\dfrac{1}{\sqrt 3}\arctan \left(\dfrac{2x-1}{\sqrt 3}\right) + C$ can be written as $\dfrac 12 \ln \left( x^2 - x +1\right)+\dfrac{1}{2} \ln \left(\dfrac 43\right)+\dfrac{1}{\sqrt 3}\arctan \left(\dfrac{2x-1}{\sqrt 3}\right) + C$ which is equal to $\dfrac 12 \ln \left( x^2 - x +1\right)+\dfrac{1}{\sqrt 3}\arctan \left(\dfrac{2x-1}{\sqrt 3}\right) + C'$.

So, this is basically it. The both answers are correct. Only their constants of integration are different.

2

$$\frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3)+C$$ Can be written as $$\frac 1 2 \log (\frac 4 3) +\frac 1 2\log( x^2-x+ 1)+C $$

Where first term can be added to the constant $$\frac 1 2 \log (\frac 4 3) +C $$ write it as new constant $C_1$

Deepesh Meena
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