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I'm currently trying to integrate: $$ \int \! \frac{2u}{u-u^3} \, du = \ln \frac{u+1}{u-1} + \ln C $$

I've tried to use partial fractions to simplify the $$ \frac{1}{u-u^3} = \frac{1}{u} - \frac{1}{2 \ln{(1+u)}} + \frac{1}{2 \ln{(1-u)}} $$ and then do integration by parts, but it doesn't look like quite right.

Can someone point me in the right direction?

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    $\begingroup$ factorise $u-u^3$ to $u(1-u)(1+u)$ and solve from there $\endgroup$
    – hwood87
    May 9, 2022 at 22:36
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    $\begingroup$ What did you do with the numerator? Please show your work so far. $\endgroup$ May 9, 2022 at 22:39
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    $\begingroup$ @Joh I did integration by parts using: - the partial fraction expression from 1/(u-u^3) -and the 2u The partial fraction of 1/(u-u^3) gave me 1/u -1/2(1+u) +1/2(1-u). However, when integrating by parts with this and 2u, a = 2u ; b = partial fraction expression a' = u^2 /2 ; b = ln(u) - ln(1+u)/2 - ln(1-u)/2 and subbing this into the formula for integration by parts does not seem like it would give the answer. $\endgroup$
    – Almond
    May 9, 2022 at 22:40
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    $\begingroup$ You do not need integration by parts. This should be done exclusively by partial fraction decomposition. You have to know how to deal with each type of partial fraction. $\endgroup$ May 9, 2022 at 22:48
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    $\begingroup$ @Almond How could a partial fraction decomposition of a rational function involve logarithms? The very first observation you should make is that $\frac{{2u}}{{u - u^3 }} = \frac{2}{{1 - u^2 }}$. $\endgroup$
    – Gary
    May 9, 2022 at 23:19

2 Answers 2

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You have : \begin{align} \frac{2u}{u-u^3} &=\frac{2u}{u(1-u^2)}\\ &=\frac{2}{(1+u)(1-u)}\\ &= \frac{2+u-u}{(1+u)(1-u)} \\ &=\frac{1}{1+u}+\frac{1}{1-u}\\&=\frac{1}{1+u}-\frac{1}{u-1} \end{align} When we integrate : $$\int \frac{2u}{u-u^3} \mathrm{d}u = \int \frac{1}{1+u}-\frac{1}{u-1} \mathrm{d}u = \ln\left(\frac{u+1}{u-1}\right)+\text{C}$$

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To calculate the indefinite integral $\int \dfrac{2u}{u-u^{3}}du$:

  1. Factorize the denominator $$ \int{\dfrac{2u}{u(1-u)(1+u)}}du $$
  2. Assume $$ \dfrac{2u}{u(1-u)(1+u)} = \dfrac{P}{u} + \dfrac{Q}{1-u}+ \dfrac{R}{1+u} $$
  3. Evaluate $P,Q,\text{ and }R$ $$ \begin{align} \dfrac{2u}{u(1-u)(1+u)} &= \dfrac{P}{u} + \dfrac{Q}{1-u}+ \dfrac{R}{1+u}\\ &=\dfrac{P(1-u)(1+u)+Q(u)(1+u)+R(u)(1-u)}{u(1-u)(1+u)}\\ &=\dfrac{u^{2}(Q-P-R)+u(Q+R)+P}{u(1-u)(1+u)}\\ \implies Q-P-R &= 0\\ Q+R &= 2\\ P &= 0\\ \implies P,Q,R = 0,1,1 \end{align} $$
  4. Substitute the values of $P,Q,\text{ and },R$ $$ \begin{align} \dfrac{2u}{u(1-u)(1+u)} &= \dfrac{0}{u} + \dfrac{1}{1-u}+ \dfrac{1}{1+u}\\ &= \dfrac{1}{1-u}+ \dfrac{1}{1+u} \end{align} $$
  5. Evaluate the integral using $\int{\dfrac{1}{a+bx}dx}=\dfrac{1}{b}\ln{(a+bx)} + C $ $$ \begin{align} &\int{\dfrac{2u}{u(1-u)(1+u)}}du\\ &=\int{\left[ \dfrac{1}{1-u}+\dfrac{1}{1+u} \right]}du\\ &=\int{\dfrac{1}{1-u}}du + \int{\dfrac{1}{1+u}}du\\ &=-\ln{(1-u)} + \ln{(1+u)} + C\\ &=\ln{\left(\dfrac{1+u}{1-u}\right)}+C \end{align} $$
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