# help with this antiderivative

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• December 6th 2009, 09:53 AM
rainer
help with this antiderivative
Where did I go wrong here...

I need to find this AD:

$\int \frac{1-ix}{1+ix}dx$

$\int \frac{1-ix}{1+ix}dx= \int \frac{i+x}{i-x}dx$

Integration by substitution...

$\int \frac{i+x}{i-x}dx= (i+x)\ln(i-x)-\int \ln(i-x)dx$

$=(i+x)\ln(i-x)-(i-x)\ln(i-x)-i+x$

$=2x\ln(i-x)-i+x$

An online integration calculator gave me:

$-2i\ln(x-i)-x$

I can't figure out where my mistake is.
• December 6th 2009, 11:18 AM
Amer
Quote:

Originally Posted by rainer
Where did I go wrong here...

I need to find this AD:

$\int \frac{1-ix}{1+ix}dx$

$\int \frac{1-ix}{1+ix}dx= \int \frac{i+x}{i-x}dx$

Integration by substitution...

$\int \frac{i+x}{i-x}dx= (i+x)\ln(i-x)-\int \ln(i-x)dx$

$=(i+x)\ln(i-x)-(i-x)\ln(i-x)-i+x$

$=2x\ln(i-x)-i+x$

An online integration calculator gave me:

$-2i\ln(x-i)-x$

I can't figure out where my mistake is.

your mistake in

$\int \frac{i+x}{i-x}dx= -(i+x)\ln(i-x)+ \int \ln(i-x)dx$

how you make the next step

you can solve it like this

$\int \frac{i+x}{i-x} dx = - \int \frac{i+x}{x-i} dx$

$-\int \frac{x-i + 2i }{x-i } dx = - \int 1 + \frac{2i}{x-i} dx$

$-\int 1 dx - \int \frac{2i}{x-i} dx = -x - 2i\ln (x-1)$