Basic Integration Problem: $\int \frac{di}{i-V/R}$

Question

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

Answer 1

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.

Answer 2

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?