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$\int \frac{x+2}{2x^3+3x^2+3x+1}\, \mathrm{d}x$

I can get it down to this:

$\int \frac{2}{2x+1} - \frac{x}{x^2+x+1}\, \mathrm{d}x $

I can solve the first part but I don't exactly follow the method in the book.

$ = \ln \vert 2x+1 \vert - \frac{1}{2}\int \frac{\left(2x+1\right) -1}{x^2+x+1}\, \mathrm{d}x $ $= \ln \vert 2x+1 \vert - \frac{1}{2} \int \frac{\mathrm{d}\left(x^2+x+1\right)}{x^2+x+1} + \frac{1}{2}\int \dfrac{\mathrm{d}x}{\left(x+\dfrac{1}{2}\right)^2 + \frac{3}{4}} $

For the 2nd part:

I tried $ u = x^2+x+1 $ and $\mathrm{d}u = 2x+1\, \mathrm{d}x$ that leaves me with $\frac{\mathrm{d}u - 1}{2} = x\, \mathrm{d}x$ which seems wrong.

because $x^2+x+1$ doesn't factor, I don't see how partial fractions again will help.

$x = Ax+B$ isn't helpful.

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    My favourite would be $x+\frac{1}{2}=w\sqrt{\frac{3}{4}}$. If you wish, I can explain it in two steps. But in the meantime, solve the problem with this substitution.2012-11-28

2 Answers 2

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The post indicates some difficulty with finding $\int \frac{dx}{x^2+x+1}$. We solve a more general problem. But I would suggest for your particular problem, you follow the steps used, instead of using the final result.

Suppose that we want to integrate $\dfrac{1}{ax^2+bx+c}$, where $ax^2+bx+c$ is always positive, or always negative. We complete the square.

In order to avoid fractions, note that equivalently we want to find $\int \frac{4a\,dx}{4a^2 x+4abx+4ac}.$ So we want to find $\int \frac{4a\,dx}{(2ax+b)^2 + (4ac-b^2)}.$ Let $2ax+b=u\sqrt{4ac-b^2}.$ Then $2a\,dx=\sqrt{4ac-b^2}\,du$. Our integral simplifies to $\frac{2}{\sqrt{4ac-b^2}}\int\frac{du}{u^2+1},$ and we are finished.

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    As I wrote in the post, try it on your problem. Start with $\frac{4}{4x^2+4x+4}=\frac{4}{(2x+1)^2+3}$, let $\sqrt{3}u=2x+1$, and you are close to the end.2012-11-30
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Let

$x+\frac12=\frac{\sqrt3}2\tan t,dx=\frac{\sqrt3}2\sec^2tdt$ $\frac12\int\frac{dx}{\left(x+\frac12\right)^2+\frac34}=\frac{\sqrt3}4\int\frac{\sec^2tdt}{\frac34\tan^2t+\frac34}=$ $\frac{\sqrt3}4\int\frac{\sec^2tdt}{\frac34\sec^2t}=\frac{\sqrt3}3\int dt=\frac{\sqrt3}3t+C$

Now just solve for $t$ in terms of $x$.

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    @MaoYiyi The idea is to use the property that $1+\tan^2\theta=\sec^2\theta$. A series of manipulations could get it into the form $k\int\frac{du}{1+u^2}=k\tan^{-1}u$. This just gets there in one step.2012-11-29