Integration using partial fraction is wrong $$\int_{2}^{\infty} \frac{1}{x-x^{3}}dx$$
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)
 A: 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 
$$\int_2^\infty \frac{1}{x}+\frac{0.5}{1-x}-\frac{0.5}{1+x}\,dx=\ln\frac{x}{\sqrt{x^2-1}}\lvert_2^\infty\ \ \text{?} $$
A: $$\int\frac{1}{x}dx=\ln{x}+C_1$$ for $x>0$ and $$\int\frac{1}{x}dx=\ln(-x)+C_2$$ for $x<0.$
In our case $x\geq2$.
Thus, $$\int_2^{+\infty}\frac{1}{x-x^3}dx=\int_2^{+\infty}\left(\frac{1}{x}-\frac{1}{2(x-1)}-\frac{1}{2(x+1)}\right)dx=$$
$$=\left(\ln{x}-\frac{1}{2}\ln(x-1)-\frac{1}{2}\ln(x+1)\right)\big{|}_2^{+\infty}=$$
$$=\ln\frac{x}{\sqrt{(x^2-1)}}\big{|}_2^{+\infty}=0-\ln\frac{2}{\sqrt3}=-\ln\frac{2}{\sqrt3}=\ln\frac{\sqrt3}{2}.$$
A: Since $u=1-x$ gives $\int f^\prime(1-x)dx=\color{blue}{-}f(1-x)+C$, $$\int_2^\infty\frac{dx}{x(1-x^2)}=\frac12\int_2^\infty\left(\frac{2}{x}+\frac{1}{1-x}-\frac{1}{1+x}\right)dx\\=\frac12\left[2\ln|x|\color{blue}{-}\ln|1-x|-\ln|1+x|\right]_2^\infty\\=\frac{-2\ln 2+\ln 3+\lim_{x\to\infty}\ln\frac{x^2}{|1-x^2|}}{2}=\ln\frac{\sqrt{3}}{2}.$$
A: 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?
