# Integrate $2u/(u-u^3)$ [closed]

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?

• factorise $u-u^3$ to $u(1-u)(1+u)$ and solve from there May 9, 2022 at 22:36
• What did you do with the numerator? Please show your work so far. May 9, 2022 at 22:39
• @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. May 9, 2022 at 22:40
• 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. May 9, 2022 at 22:48
• @Almond How could a partial fraction decomposition of a rational function involve logarithms? The very first observation you should make is that $\frac{{2u}}{{u - u^3 }} = \frac{2}{{1 - u^2 }}$.
– Gary
May 9, 2022 at 23:19

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 : $$\int \frac{2u}{u-u^3} \mathrm{d}u = \int \frac{1}{1+u}-\frac{1}{u-1} \mathrm{d}u = \ln\left(\frac{u+1}{u-1}\right)+\text{C}$$
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}