Mathematics

Solve:
$$\int {\dfrac{x}{{{x^2} + x + 1}}} dx$$


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

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

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

$$ \displaystyle = \frac{1}{2}log (x^{2}+x+1)-\frac{1}{2}\int \frac{1}{x^{2}+2.x\frac{1}{2}+(\frac{1}{2})^{2}-(\frac{1}{2})^{2}+1} $$

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

Let $$\displaystyle x+\frac{1}{2} = t \Rightarrow dx = dt $$

$$ \displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{2}\int \frac{dt}{t^{2}+(\frac{\sqrt{3}}{2})^{2}} $$

$$ \displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{2}.\frac{1}{(\frac{\sqrt{3}}{2})}ten^{-1}(\frac{t}{(\sqrt{3}/2)})+c $$

$$ \displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{\sqrt{3}}tan^{-1}(\frac{2(x+1/2)}{\sqrt{3}})+c $$

$$ \displaystyle = \frac{1}{2}log(x^{2}+x+1)-\frac{1}{\sqrt{3}}tan^{-1}(\frac{2x+1}{\sqrt{3}})+c $$

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Subjective Medium Published on 17th 09, 2020
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