Basic Integration Problem: $\int \frac{di}{i-V/R}$ This is my first time using the maths stack forum, I normally use the electrical engineering stack. I am having difficulty taking the integral of a term (see the following link).
$$\int \frac{di}{i}=\ln(i)+C \tag{1}$$
$$\int \frac{di}{i-V/R}=\ln\left(\frac{i}{?}\right)+C \tag{2}$$
I understand equation $(1)$ which the integral is just the natural log of $i$.
But with equation $(2)$, I don't know how to integrate this due to the $-V/R$ on the bottom line. How is this done with direct integration (not using substitution)?
Thanks
 A: If one does allow substitution, we may write
$j = i - \dfrac{V}{R}; \tag 1$
then
$dj = di, \tag 2$
whence
$\displaystyle \int \dfrac{di}{i - V/R} = \int \dfrac{dj}{j} = \ln j + C = \ln \left ( i - \dfrac{V}{R} \right) + C ; \tag 3$
at this point we may check, via differentiation, using the chain rule, but without substitution:
$\left (\ln \left ( i - \dfrac{V}{R} \right) + C \right )' = \left (\ln \left ( i - \dfrac{V}{R} \right) \right )'$
$= \dfrac{1}{i - V/R} \dfrac{d(i - V/R)}{di} = \dfrac{1}{i - V/R} \dfrac{di}{di} = \dfrac{1}{i - V/R}. \tag 4$
Of course, once the derivative in (4) is computed, we have at our disposal an anti-derivative or primitive for $1/(i - V/R)$, namely $\ln(i - V/R) + C$, so now the fundamental theorem of calculus directly yields
$\displaystyle \int \dfrac{di}{i - V/R} = \ln \left ( i - \dfrac{V}{R} \right) + C,  \tag 6$
without introducing an auxiliary variable such as $j$.
Synopsis: To do it with direct integration, know and use the primitive of $1/(1 - V/R)$ given in (4).  The End.
A: Hint:
Set $a=V/R$ and substitute $\begin{bmatrix}t \\ \mathrm dt \end{bmatrix}=\begin{bmatrix} i-a \\ \mathrm di\end{bmatrix}\implies \displaystyle\int\dfrac{\mathrm di}{i-V/R}=\displaystyle\int \dfrac{\mathrm dt}{t}$. Can you proceed?
