Mathematics

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


SOLUTION
I=$$\int\dfrac{x}{x^2+x+1}dx=\int\dfrac{\dfrac{1}{2}(2x+1)-\dfrac{1}{2}}{x^2+x+1}dx=\dfrac{1}{2}\int\dfrac{2x+1}{x^2+x+1}dx(i_{1})-\dfrac{1}{2}\int\dfrac{1}{x^2+x+1}dx(i_{2})+C$$

$$i_{1}$$ can be read as $$\dfrac{1}{2}\int\dfrac{dt}{t}$$ for t=$$x^2+x+1$$
so $$i_{1}=\dfrac{1}{2}ln|x^2+x+1|+c$$

$$i_{2}$$ can simply be evaluated  by making perfect square in denominator 
$$i_{2}=\dfrac{1}{2}\int\dfrac{1}{(x+\dfrac{1}{2})^{2}+\dfrac{3}{4}}=\dfrac{1}{\sqrt{3}}tan^{-1}(\dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}})+c$$

So I=$$i_{1}-i_{2}=$$ $$\dfrac{1}{2}ln|x^2+x+1|-\dfrac{1}{\sqrt{3}}tan^{-1}(\dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}})+C$$

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Subjective Medium Published on 17th 09, 2020
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