\[ \large \int \dfrac{1}{x^2+1} \, dx = arctanx + k \] but the same integral can be written as

\[ \large \int \dfrac{1}{x^2-i^2} \, dx \] where\[ i=\sqrt{-1} \]
Using partial fractions and then integrating
\[ \large \int \dfrac{1}{x^2-i^2} \, dx = 1/2i* \ln \frac { x-i }{ x+i } + C \]
Prove that their difference is a constant.

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x - i =re^(iy)

So taking conjugates both sides,

x + i = re^(-iy)

So,

ln (x-i) - ln (x+i)= ln (re^(iy)) - ln (re^(-iy))= 2iy

Here y= arctan (-1/x) = arctan (x) - pi/2 .

So the integrals differ by a constant. Hope this helps sorry for typing errors. – Abhi Kumbale · 9 months, 2 weeks ago

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You need to know that log of imaginary number is defined – Dev Rajyaguru · 10 months ago

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– Priyamvad Tripathi · 10 months ago

Yes I came to know it after I posted this note and even proved the same.Log in to reply

Do you mean \[\frac { x-i }{ x+i } \] or \[x-\frac { i }{ x+i } \] ? and \[\frac{1}{2}i\] or \[\frac{1}{2i}\]? – Hummus A · 10 months, 1 week ago

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– Priyamvad Tripathi · 10 months, 1 week ago

Sir I have corrected it. It was the first one \[\frac { x-i }{ x+i } \]Log in to reply