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$$\int_{2}^{\infty} \frac{1}{x-x^{3}}dx$$

I think the way to solve this, is using partial fraction. For some reason I can't get to the answer...

$$\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}=\frac{1}{x-x^{3}}$$

$$A(1-x)(1+x)+Bx(1+x)+Cx(1-x)=1$$

$$x=1: 2B=1=>B=0.5$$

$$x=-1:-2C=1=>C=-0.5$$

$$x=0: A=1$$

Then rewriting the integral:

$$\int \frac{1}{x}+\frac{0.5}{1-x}-\frac{0.5}{1+x} $$

$$\ln x+0.5\ln(1-x)-0.5\ln(1+x)|_{2}^{\infty}$$

Inserting 0.5 into the brackets and using log rules I get:

$$\ln x+\ln(\sqrt\frac{1-x}{1+x})$$

$$x \to \infty $$ inside the square root the result is -1 as 'X' approaches infinity, and I can't sqrt of a negative number.

Perhaps I was wrong somewhere along the lines, maybe my way of integration is wrong...

Answer: 0.5*ln(3/4)

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Hint: The antiderivative of $\frac{1}{1-x}$ is not $\ln(1-x)$. Actually, the antiderivative of $\frac1x$ is $\ln|x|+C$, in the sense that $\int\frac1x=\ln x+C$ for $x>0$ and $\int\frac1x=\ln (-x)+C$ for $x<0$.

Can you use this to obtain $$\int_2^\infty \frac{1}{x}+\frac{0.5}{1-x}-\frac{0.5}{1+x}\,dx=\ln\frac{x}{\sqrt{x^2-1}}\lvert_2^\infty\ \ \text{?} $$

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    $\begingroup$ Your statement that $\int\frac{1}{x}dx=\ln|x|+C$ is wrong. The right statement is: $\int\frac{1}{x}dx=\ln{x}+C_1$ for $x>0$ and $\int\frac{1}{x}dx=\ln(-x)+C_2$ for $x<0.$ $\endgroup$ – Michael Rozenberg Jul 20 at 9:16
  • $\begingroup$ I see how you reached the final exp. but I cant see how placing actual values to x gets me to the answer. $\endgroup$ – user6394019 Jul 20 at 9:19
  • $\begingroup$ @MichaelRozenberg Sorry for the ambiguous statement. Fixed now. $\endgroup$ – Feng Shao Jul 20 at 9:21
  • $\begingroup$ @user6394019 $\ln\frac{x}{\sqrt{x^2-1}}\lvert_2^\infty=0-\ln\frac{2}{\sqrt 3}=\ln\frac{\sqrt 3}{2}=0.5\ln(3/4).$ $\endgroup$ – Feng Shao Jul 20 at 9:23
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$$\int\frac{1}{x}dx=\ln{x}+C_1$$ for $x>0$ and $$\int\frac{1}{x}dx=\ln(-x)+C_2$$ for $x<0.$

In our case $x\geq2$.

Thus, $$\int_2^{+\infty}\frac{1}{x-x^3}dx=\int_2^{+\infty}\left(\frac{1}{x}-\frac{1}{2(x-1)}-\frac{1}{2(x+1)}\right)dx=$$ $$=\left(\ln{x}-\frac{1}{2}\ln(x-1)-\frac{1}{2}\ln(x+1)\right)\big{|}_2^{+\infty}=$$ $$=\ln\frac{x}{\sqrt{(x^2-1)}}\big{|}_2^{+\infty}=0-\ln\frac{2}{\sqrt3}=-\ln\frac{2}{\sqrt3}=\ln\frac{\sqrt3}{2}.$$

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Since $u=1-x$ gives $\int f^\prime(1-x)dx=\color{blue}{-}f(1-x)+C$, $$\int_2^\infty\frac{dx}{x(1-x^2)}=\frac12\int_2^\infty\left(\frac{2}{x}+\frac{1}{1-x}-\frac{1}{1+x}\right)dx\\=\frac12\left[2\ln|x|\color{blue}{-}\ln|1-x|-\ln|1+x|\right]_2^\infty\\=\frac{-2\ln 2+\ln 3+\lim_{x\to\infty}\ln\frac{x^2}{|1-x^2|}}{2}=\ln\frac{\sqrt{3}}{2}.$$

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There is no need to do partial fractions on this one. Simply set $\frac{1}{t}=x$. Your new integral will be of the form $\frac{t}{t^2-1}$ and that's simply a natural log. That's it! Can you finish from here?

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