Evaluating $\int \frac{L}{E-iR}\operatorname d i$

Question

Can someone please explain to me step by step, how to evaluate this integral? $E$, $R$ and $L$ are constants.

$$\int \frac{L}{E-iR}di$$

The result should be: $-\frac{L}{R}\ln(E-iR) + C$

Thank you.

Answer 1

let $ u = E - iR$, then $\dfrac{du}{di} = -R \rightarrow du = - R \,di$

Then

$$ \int \frac{L}{E - iR} \ di = \frac{L}{-R}\displaystyle\int \frac{-R}{E - iR} \ di $$ $$= \frac{L}{-R} \int \frac{1}{u} \ du = \frac{L}{-R} (\ln (u) + C)$$ $$ = \frac{L}{-R} (\ln (E - iR) + C) =\frac{L}{-R} (\ln (E - iR) + K $$