okay let's do some math for fun and today we'll be integrating 1/(x^2+1) i.e. integral of 1/(1+x^2)
i know i know i know you know and you should know i know too the answer is just tan^-1(x)
and of course don't forget the plus c and that's it no of course not because otherwise you have so
i'm going to unsubscribe right so let's go ahead and see how to do this legitimately have a look
on the button here we have x^2+1 let's factor it we get (x-i)(x+i) this is legitimate all right and
then it's also legitimate to do partial fractions even with complex numbers so here we have x minus
i plus x plus i and then dx and to figure this and that that's just constants it's still legitimate
to do partial fractions with the kapha method as well to figure this out we go back to the original
and cover this denominator and we have to ask ourselves how can we make x minus i equals zero
x has to be i and we put this side right here i plus i is 2y on the bottom so we get 1 over 2i
and then to figure this out we go back to the original cover the same denominator
and ask ourselves how can we make x plus i equals 0 x has to be negative i and we put that right
here and we'll end up with 1 over negative 2i and now let's factor out the constant so we have 2i
on the bottom and also the one and then put on parenthesis and now we just have to integrate
1 over x to the first power which will give us natural log not only for absolute value we're
talking about imagine numbers already so just put on parentheses and then we have the x minus i
and here we have a minus and then secondly we also have the ln x plus i cool plus c done ready
ready does this look like what we have right here of course it looks so different right
and also perhaps the biggest question is that is this even real look we have i i i
but you know for sure this expression has to be real because originally this integral is in
the real world so now we just have to see how we can make sense of this and make sense of that and
hopefully you can cancel things out and we get back to the real world so let's get to work on
that um the truth is when we have a complex number instead of the natural log we can simplify it and
to do so we will have to change this to its polar form and let me just put this down right here
for you guys let's look at the polar plane that's that we have a complex number here a plus b i this
right here can be written in the polar form and that will be r e i theta 1 so why the r theta well
r is the distance from the origin to the point and then three times the angle from here to here and
the truth is there are infinitely many angles that will satisfy this because for example once you
have here you can keep adding two pi or subtract 2 pi and you still end up with the same place
but we'll just be looking at the principal branch it's kind of like saying the first answer
and that will actually give us the answer for this like the real answer
all right so now let's see we know the horizontal distance is the a and then the vertical distance
is the b because this is the real axis this is the imaginary axis so right here we can
say r thanks to the passive equivalent theorem is the square root of a squared plus b squared
and then for the angle theta we can use tangent and we can say tangent
theta equals opposite over adjacent or we can say theta is equal to inverse tangent
of opposite over adjacent aha we have a good hope of coming up with the inverse tangent right cool
now let's see what's x minus i did well remember x is real okay and then we have the negative 1 i so
right here i'm just going to put it down for you guys here we still have that 1 over 2i and then we
have the ln and this right here the r will be this so we have the square root let me just put this
down right here we have the square root and then x squared and then minus 1 squared right so i'll
just put that down minus 1 squared like this plus negative 1 square technically and then we have the
e to the i and then theta is inverse tangent and then b is negative 1 and a is x so we have
negative 1 over x cool and then we are going to subtract the second part so minus ln and then
here it's pretty much the same thing but b here is positive one so we have square root x squared
plus one square and then times e to the i theta is inverse tangent and then we have positive one
over x cool all right now the good thing is that whenever we have the ln of a product we can just
break it apart as ln of this plus ln of that so let's just go ahead and write it down and
then just be really careful this is a subtraction that's why it becomes subtraction and subtraction
now check this out this and that cancel each other out and then when we have e sub d ln of course
they also cancel each other out and again we're just looking for the principal value all right
so we don't add the 2n pi and now for this power we just have i times inverse tangent of
negative 1 over x but inverse tangent is an odd function so for this negative we can put it on
outside so this right here will give us 1 over 2 i and then we have this as negative i
times inverse tangent of 1 over x and the truth is once we cancel this and that this is also a
minus i inverse tangent of 1 over x so together we have minus 2i and then inverse tangent
of 1 over x very nice and now ladies and gentlemen 2i 2i cancelled nice d so on o we just get
negative inverse tangent of 1 over x and then we are done no of course i didn't put on the plus c
but now the question is is this really the same as inverse tangent no in fact this is
not exactly the same as inverse tangent the truth is we will have to use the green marker
this right here we have a very nice identity when we have inverse tangent of x if we add it with
inverse cotangent of x this right here becomes pi over 2 which is very nice but i'm not using
inverse cotangent i'm using that the truth is this right here is in fact inverse tangent of 1 over x
now check this out this is positive we can just move it to the other side so we have the negative
and then we can put the pi over 2 to the other side so in fact we can look at this as inverse
tangent of x and then minus pi over 2 and don't forget to put down the plus c
and technically right here we should put down a c1 and c1 because of course it's like we have
this additional negative power too but the truth is this is a constant that's also a constant
together they're still just another constant so we can label this as c2 but you know what
we're done with the equation so let's not worry about the subscript so ta-da
all this root root and we get inverse tangent of x and then plus c and then we are done
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