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$$\int \frac{dx}{x^3+1}$$

so at first $\frac{1}{x^3+1}=\frac{-x+2}{3(x^2-x+1)}+\frac{1}{3(x+1)}$

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

For

$$\frac{1}{3}\int\frac{dx}{(x+1)}=\frac{1}{3}ln|x+1|+C$$

For

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

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

For

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

How should I approach $$-\frac{1}{3}\int \frac{x}{(x-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}dx$$?

3SAT
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gbox
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    Let's start from $\int \frac{-x+2}{x^2-x+1}dx$ (I did drop the constant in front of it, you can always add it in later). What you should do is to differentiate the denominator, so: $(x^2-x+1)' = 2x-1$, and we'll try to split the integral to something like this: $... = \int \frac{-\frac{1}{2}(2x-1)+\frac{3}{2}}{x^2-x+1}dx = -\frac{1}{2}\int \frac{2x-1}{x^2-x+1}dx + \frac{3}{2} \int \frac{1}{x^2-x+1} dx $, can you go from here? Hint, the first integral needs a simple substitution, and the second one, I think you can handle it. – user49685 Feb 16 '16 at 18:10

3 Answers3

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$$\int\frac{-x+2}{x^2-x+1}dx=-\frac12\int\frac{2x-1}{x^2-x+1}dx+\frac32\int\frac{dx}{\frac34+\left(x-\frac12\right)^2}$$

$$=-\frac12\log(x^2-x+1)+\sqrt3\int\frac{\frac2{\sqrt3}dx}{1+\left(\frac{2x-1}{\sqrt3}\right)}=\log\frac1{\sqrt{x^2-x+1}}+\sqrt3\arctan\frac{2x-1}{\sqrt3}+C$$

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

Set $u=x-\frac 1 2$ and $du=dx$

$$=-\frac 1 3 \int \frac{u+1/2}{u^2+3/4}du=-\frac 1 3 \int \frac {u}{u^2+3/4}du-\frac 1 6 \int \frac{du}{u^2+3/4}$$

Set $s=u^2+\frac 3 4 $ and $ds=2udu$, set $p=\frac{2u}{\sqrt 3}$ and $dp=\frac{2}{\sqrt 3}du$

$$=-\frac 1 6 \int \frac 1 s-\frac{1}{3\sqrt 3}\int \frac{dp}{p^2+1}=\boxed{-\frac 1 6 \ln|x^2-x+1|-\frac{\arctan(\frac{2x-1}{\sqrt 3})}{3\sqrt 3}+C}$$

3SAT
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I thought it might be instructive to present a general methodology.

In THIS ANSWER, I showed that the integral $\int \frac{1}{1+x^n}\,dx$ can be expressed as

$$\bbox[5px,border:2px solid #C0A000]{\int\frac{1}{x^n+1}dx=-\frac1n\sum_{k=1}^n\left(\frac12 x_{kr}\log(x^2-2x_{kr}x+1)-x_{ki}\arctan\left(\frac{x-x_{kr}}{x_{ki}}\right)\right)+C} \tag 1 $$

where $x_{kr}$ and $x_{ki}$ are the real and imaginary parts of $x_k$, respectively, and are given by

$$x_{kr}=\text{Re}\left(x_k\right)=\cos \left(\frac{(2k-1)\pi}{n}\right) \tag 2$$

$$x_{ki}=\text{Im}\left(x_k\right)=\sin \left(\frac{(2k-1)\pi}{n}\right) \tag 3$$

For $n=3$, we can use $(1)-(3)$ and arrive at

$$\bbox[5px,border:2px solid #C0A000]{\int\frac{1}{x^3+1}dx=\frac13 \log|x+1|-\frac16 \log(x^2-x+1)+\frac{\sqrt3}{3}\arctan\left(\frac{2x-1}{\sqrt3}\right)+C}$$

Community
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Mark Viola
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