when we see the integral of 1/(1+x^2) of course we know the answer for this is just equal to
tan^-1(x)+C and if you want to show work of course we can just do trig sub let x equal to
tangent Theta and then we can continue from here and this is usually done in calculus 2 but last
time also showed you guys how we can use complex numbers to do this and that is to factor 1 plus
x squared as I plus x times negative I plus X and then two partial fractions in the complex
world but I did that for you guys last time already so if you haven't seen that video go
ahead and watch that after this video the link will be in the description for your convenience
so what exactly are we doing here today well don't worry I'm going to square the denominator
easy right because we can just Square this no of course not
don't do that the question is can we still do tricks up the answer for this yes
can we do this also yes but in fact we are not going to do this anymore instead I will show you
guys a very powerful technique for integration and that's called the finance technique of integration
AKA differentiation under the integral sign AKA that means true for integration so how exactly
do we do it though well first we have to start with something that we know so check this out
but what if this number is now one let's say it's 17. well in that case what do we do on the right
hand side right so as you can see this formula is not so complete yet but instead this formula is
better that is when we have a a squared right here when we have that the right hand side is
going to be 1 over a times the inverse tangent of X over a and you can do tricks up for this as well
so I'm going to delete that to you guys I'm just going to use this formula to show you this a is
called the parameter because it kind of generates like the old complete version of this formula
for you guys right so how exactly can we utilize this though five minutes is going to tell us that
whenever you have a formula with parameter just go ahead and differentiate that with respect to the
parameter and hope for the good things to happen and it will happen trust me he didn't say that but
I made up so let's go ahead and check it out I'm just going to differentiate this with respect to a
like so well on the left hand side we are differentiating an integral so what do we do well
this is the time that we have to use the so-called differentiation on the dental cosine that tells us
that we can bring this inside but change the D to the partial derivative symbol because we have the
A and also the X inside here so this definite side becomes the integral and then partial derivative
and right here to differentiate that let's just go ahead and write this
as a square plus x square raised to a negative one and we still have the DX
on the right hand side I'm just going to keep yes take the derivative and we have this right here
so it's the one over a but this right here will be our first function because a is the variable
here and then times the inverse tangent of X over a and of course for the plastic right here
Okay so let's go ahead and work out the derivative for this use the power rule first so bring
the power to the front and then minus one so we have negative a squared plus x squared raised to
the negative two but don't forget the chain rule multiply by derivative inside with respect to a so
that is just going to be too high the derivative x squared is zero because we are in the a world
so that's the derivative from the inside but we still have the integral so go ahead put down the
integral right here and see some of you guys see what we are doing on the right hand side
if we can figure out the derivative we pretty much will have the formula well you have to use
the product rule we are going to keep the first function which is 1 over a times the derivative
inverse tangent it will give us 1 over 1 plus whatever this is a square so X over a square
that's not yellow we will have to use the chain rule so X is just a constant multiple
in the a world so let's put that down and then differentiating one over eight write that as a
to the negative one will get Negative a to the negative two so it's just negative one over a
squared good and then we will continue keep the second function so plus inverse tangent of X over
a times Theta derivative of that which is negative one over a squared derivative C is just zero
can we differentiate both sides with respect to C don't do that you could but why
yeah anyway though that's pretty much it now we just have to clean things up so on the left hand
side let me just put this down as the integral negative 2A over and then here already as a
squared plus x squared and then square and then d a this right here to clean this up okay we have
investable parentheses on the bottom here I'm just going to take the a square multiplying multiplying
on the top we have negative 1 times x so that's Negative X over this a is just chilling at the
front right here this time start is a squared this times that is x squared so that's the first part
and for the second part we'll just say minus 1 over a squared times inverse tangent of X over a
yeah we're pretty much done but I would like to divide Everybody by
2A here so that it's cleaner Jose so let's go ahead oh sorry it should be a negative
right here negative 2A so let's divide the value by negative 2A so 1 over negative 2A
so finally integral 1 over a squared plus x squared and then Square DX no partial fraction no
tricks up yeah differentiation under the interco side give us that negative negative
becomes positive we have just X on the top over 2 a squared times a squared plus x squared and then
negative negative becomes positive 1 over 2 a to the third power and then inverse tangent of X over
a wow how cool is this right and because we are talking about formula and this is the intervene
integral so at the end here I will put down plus C so ladies and gentlemen this is a very
nice formula for you guys with this approach and of course we should finish this question right
so this and that is just when the case a is equal to 1. can you put negative one yes
ah interesting but why don't give us negative negatives and passes so one's okay so I'm just
going to say that a equal to 1. so that's that and then on the right hand side we'll just have
X over 2 and then 1 plus x squared and then right here just one over two and we have inverse tangent
of X over a which is just over one so we have this and finally put a plus c just like that so
what do you guys think go up hey I hope you liked the video so far and if you enjoying
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