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

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



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


$$x=0: A=1$$

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)


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 $$\int_2^\infty \frac{1}{x}+\frac{0.5}{1-x}-\frac{0.5}{1+x}\,dx=\ln\frac{x}{\sqrt{x^2-1}}\lvert_2^\infty\ \ \text{?} $$

  • 1
    $\begingroup$ 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

$$\int\frac{1}{x}dx=\ln{x}+C_1$$ for $x>0$ and $$\int\frac{1}{x}dx=\ln(-x)+C_2$$ for $x<0.$

In our case $x\geq2$.

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


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


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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.