Janet Forbes

2022-07-04

I need to compute the integral $\int \frac{2x}{\left({x}^{2}+x+1{\right)}^{2}}\cdot dx$. I tried using the integration of a rational function technique, with $\frac{Ax+B}{{x}^{2}+x+1}+\frac{Cx+D}{\left({x}^{2}+x+1{\right)}^{2}}$, but this simply returned $C=2$ and $A,B,D=0$, so it doesn't really change anything.
I also tried using a $u$ substitution, setting $u={x}^{2}+x+1$. This made the numerator $2x=\frac{du}{dx}-1$, but I'm not really sure if I can do that/how to solve an integral with a derivative as a part of it.
How would I go about solving this?

1s1zubv

Expert

$2\int \frac{x}{\left({x}^{2}+x+1{\right)}^{2}}$
$2\int \frac{x}{\left(\left(x+\frac{1}{2}{\right)}^{2}+\frac{3}{4}{\right)}^{2}}$
Apply u-substitution: $u=x+\frac{1}{2}$
$2\int \frac{8\left(2u-1\right)}{\left(4{u}^{2}+3{\right)}^{2}}du$
$2\left(8\right)\int \frac{2u-1}{\left(4{u}^{2}+3{\right)}^{2}}du$
Apply the Sum Rule
$2\left(8\right)\left(\int \frac{2u}{\left(4{u}^{2}+3{\right)}^{2}}du-\int \frac{1}{\left(4{u}^{2}+3{\right)}^{2}}du\right)$
Now,
$\int \frac{2u}{\left(4{u}^{2}+3{\right)}^{2}}=-\frac{1}{4\left(4{u}^{2}+3\right)}$
Now,
$\int \frac{1}{\left(4{u}^{2}+3{\right)}^{2}}du=\frac{1}{12\sqrt{3}}\left(arctan\left(\frac{2}{\sqrt{3}}u\right)+\frac{1}{2}sin\left(2arctan\left(\frac{2}{\sqrt{3}}u\right)\right)\right)$
$=2\left(8\right)\left(-\frac{1}{4\left(4{u}^{2}+3\right)}-\frac{1}{12\sqrt{3}}\left(arctan\left(\frac{2}{\sqrt{3}}u\right)+\frac{1}{2}sin\left(2arctan\left(\frac{2}{\sqrt{3}}u\right)\right)\right)$
After doing small calculations and substituting $u=x+\frac{1}{2}$,
$\int \frac{2x}{{x}^{2}+x+1}dx=16\left(-\frac{1}{4\left(4{x}^{2}+4x+4\right)}-\frac{1}{24\sqrt{3}}\left(2arctan\left(\frac{2x+1}{\sqrt{3}}\right)+sin\left(2arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right)\right)\right)+C$

civilnogwu

Expert

Apply integration by parts to
$\int \frac{1}{{x}^{2}+x+1}dx$
and extract your desired integral from the result.

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