The (Tsiolkovsky) rocket equation states that the velocity of a rocket can be calculated as
$$v(t) = v_0 \ln\frac{m_0}{m_0 - \dot m t}$$
where $m_0$ is the starting mass, $\dot m$ is the (constant) loss of mass per time due to the rocket engine which emits the exhaust gas with the velocity $v_0$. To get the distance travelled at a certain time after the launch, I have to integrate this formula. My approach is as follows:
$$ \begin{align} &\int \ln \frac{a}{a-bt}\text{d}t\\ = &\int \ln a \text{d}t - \int \ln (a-bt)\text{d}t\\ = & t\ln a - \left( -\frac{1}{b}((a-bt) \ln(a-bt) - (a-bt)\right) \end{align} $$
I know that there is a "nice" solution: $$\int \ln \frac{a}{a-bt}\text{d}t = \frac{a-bt}{b} \ln \frac{a-bt}{a} + t$$ but I can't see how to get there from my approach. Which steps are neccessary, or is there a different way to begin with?
PS: it seems to work if I remove the constant $\frac{a}{b}$ and add the constant term $-\frac{a}{b}\ln a$ in my antiderivative. But there has to be a better way!
You may just use an integration by parts. $$ \begin{align} \int \ln \frac{a}{a-bt}\:\text{d}t&=t\:\ln \frac{a}{a-bt}-\int t\:\frac{\left(\frac{a}{a-bt}\right)'}{\frac{a}{a-bt}}\:\text{d}t\\\\ &=t\:\ln \frac{a}{a-bt}-\int t\:\frac{b}{a-bt}\:\text{d}t\\\\ &=t\:\ln \frac{a}{a-bt}+\int \frac{a-bt-a}{a-bt}\:\text{d}t\\\\ &=t\:\ln \frac{a}{a-bt}+t-a\int \frac{1}{a-bt}\:\text{d}t\\\\ &=t\:\ln \frac{a}{a-bt}+t-\frac ab\int \frac{b}{a-bt}\:\text{d}t\\\\ &=t\:\ln \frac{a}{a-bt}+t-\frac ab\ln \frac{a}{a-bt}\\\\ &=\left(t-\frac ab\right)\ln \frac{a}{a-bt}+t\\\\ &=\frac{a-bt}{b} \ln \frac{a-bt}{a} + t. \end{align} $$