Evaluating $\int_{0}^{2} \frac{t}{t+i} dt$ $$\int_{0}^{2} \frac{t}{t+i} dt$$
I have no idea how to even begin to split this up into real and imaginary parts.  The only thing I can think of is that it might involve the use of natural log? Any help on how to begin would be VERY much appreciated!
 A: You can multiply numerator and denominator by $t-i$.  That will give you a real denominator, so the real/imaginary split will be easy.
A: Splitting problems into real and imaginary parts usually makes them harder, so usually don't want to do it, unless you are specifically looking for solutions expressed, e.g., in terms of $\arctan$ rather than in terms of $\log$ and are willing to do the extra work.
The simplest evaluation method for this integral is essentially no different than the one you would use to evaluate
$$ \int_{0}^{2} \frac{t}{t-3} \, dt $$
and would involve using logarithms, as you suggested.
A: For any complex number $a$,
and ignoring any questions about meaning,
$\int \frac{t}{t+a} dt
=\int \frac{t+a-a}{t+a} dt
=\int \frac{t+a}{t+a} dt-\int \frac{a}{t+a} dt
=\int  dt-a\int \frac{1}{t+a} dt
= x - a \ln(x+a)
$.
For your case,
the result is
$(x - i \ln(x+i)) \big|_0^2
=2 - i \ln(2+i) +i \ln(i)
= 2 - i\ln(2/i+1)
=2 - i\ln(1-2i)
$.
A: You may take the imaginary part as a vertical axis, and $t$ as real axis, then integrate as if the integrand were a real division.
