Computing the integral of a rational function

Question

I need to compute the integral $\int \dfrac{2x}{(x^2+x+1)^2} \cdot dx$. I tried using the integration of a rational function technique, with $\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}$, 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?

Thanks for your time.

Answer 1

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

Answer 2

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

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

Apply u-substitution: $u=x+\frac12$

$$2\int { \frac { 8(2u-1) }{ (4u^ 2+3)^ 2 } } du$$

$$2(8)\int { \frac { 2u-1 }{ (4u^ 2+3)^ 2 } } du$$ Apply the Sum Rule $$2(8)(\int { \frac { 2u }{ (4u^ 2+3)^ 2 } } du-\int { \frac { 1 }{ (4u^ 2+3)^ 2 } du } )$$

Now, $$\int { \frac { 2u }{ (4u^ 2+3)^ 2 } } =- { \frac { 1 }{ 4(4u^ 2+3) } } $$

Now, $$\int { \frac { 1 }{ (4u^ 2+3)^ 2 } du=\frac { 1 }{ 12\sqrt { 3 } } (arctan(\frac { 2 }{ \sqrt { 3 } } u)+\frac { 1 }{ 2 } sin(2arctan(\frac { 2 }{ \sqrt { 3 } } u))) } $$

$$=2(8)(- { \frac { 1 }{ 4(4u^ 2+3) } }- { \frac { 1 }{ 12\sqrt { 3 } } (arctan(\frac { 2 }{ \sqrt { 3 } } u)+\frac { 1 }{ 2 } sin(2arctan(\frac { 2 }{ \sqrt { 3 } } u))) } $$

After doing small calculations and substituting $u=x+\frac12$,

$$\int\frac{2x}{x^2+x+1}dx=16(-\frac{1}{4(4x^2+4x+4)}-\frac{1}{24\sqrt{3}}(2arctan(\frac{2x+1}{\sqrt{3}})+sin(2arctan(\frac{2x+1}{\sqrt{3}}))))+C$$