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

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.

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Have you learned about substitution? Try $iR = Eu$ –  N3buchadnezzar Jul 28 '13 at 12:54
@rubick, try putting $E-iR=u$ –  lab bhattacharjee Jul 28 '13 at 12:57
I tried it, but now I realized I made error. I was integrating $u$ instead of doing $\frac{du}{di}$ –  rubick Jul 28 '13 at 13:08

## 1 Answer

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

Then

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

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