a question about an integral

wildrabbitt · 2019-02-03, 21:06

I was wondering what happens when the method of partial fractions is used for indefinite integrals I've always done for a real variable.

The integrand was

$\frac{1}{1+x^2}$

the solution I got was

$\frac{i}{2}\frac{ln(1 + ix)}{ln(1 - ix)} + c$

If there was nothing wrong with the method I used then the latter expression
is arctan(x) + c.

Was there anything wrong with my method?

Should have written

$\frac{i}{2}ln{\frac{(1 + ix)}{(1 - ix)} + c$

Dr Sardonicus · 2019-02-04, 20:56

Suggest you use the formula

$e^{i\theta}\;=\;\cos{(\theta)}\;+\;i\sin{(\theta)}$

2019-02-03, 21:06	#1
wildrabbitt Jul 2014 677₈ Posts	a question about an integral I was wondering what happens when the method of partial fractions is used for indefinite integrals I've always done for a real variable. The integrand was $\frac{1}{1+x^2}$ the solution I got was $\frac{i}{2}\frac{ln(1 + ix)}{ln(1 - ix)} + c$ If there was nothing wrong with the method I used then the latter expression is arctan(x) + c. Was there anything wrong with my method? Should have written $\frac{i}{2}ln{\frac{(1 + ix)}{(1 - ix)} + c$ Last fiddled with by wildrabbitt on 2019-02-03 at 21:57

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2019-02-04, 20:56	#2
Dr Sardonicus Feb 2017 Nowhere 4643₁₀ Posts	Suggest you use the formula $e^{i\theta}\;=\;\cos{(\theta)}\;+\;i\sin{(\theta)}$