Integrate $2u/(u-u^3)$

Question

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?

Answer 1

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 $ $$ \begin{align} &\int{\dfrac{2u}{u(1-u)(1+u)}}du\\ &=\int{\left[ \dfrac{1}{1-u}+\dfrac{1}{1+u} \right]}du\\ &=\int{\dfrac{1}{1-u}}du + \int{\dfrac{1}{1+u}}du\\ &=-\ln{(1-u)} + \ln{(1+u)} + C\\ &=\ln{\left(\dfrac{1+u}{1-u}\right)}+C \end{align} $$

Answer 2

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