day one okay if you can integrate this today then you will be good for the day yeah let's go ahead
and add one to here and then we guess three and then divide it by the new power so we have 1/3 so
the answer is 1/3 x to the 3 power and then don't forget the plus C that's it day two today's topic
is called the use substitution and here's the example we are going to integrate x * e to the
x² power how do we do it pay attention to to the inner function we have x s put that to be U and
then differentiate both sides we will have du = 2x DX and we see that we have X DX right here
but we need to have the two so that we can get the DU so that's come here and multiply by two
but of course we'll have to divide that right way as well so that we don't change anything yeah and
we will just get 12 integral the 2x DX X is the DU we can put that right here and then we have e
and the x squ is the U so we're just integrating e to the U in the U world and the answer for that
is just e to the U and then of course we still have that 1/2 right here and then e to the u u
is x² done huh after we put on the plus C that's it day three today's topic is called integration
by parts and we use the DI format for it this is the integral of the day we are integrating
x² * e to DX so how do we do it we'll put on the D and I first d stands for differentiating and I
stands for integrating and then also on the side just put down some plus or minus just to get ready
we will differentiate X squ because if we keep going you see 2X and then two and then zero if
you see Zero in the D column you can stop and then we just integrate e to d x you can integrate this
as many times as you want it's always going to be e to DX and do this this and that that'll pretty
much give you the answer so first we get POS x^2 * e to X and then - 2x * e to X and lastly + 2 * e
to the X and we're done after the plus c day four today's topic is called trick up and we are going
to integrate square root of 1 + x^2 and the idea is to utilize sound trick identities to help us
simplify this because it has two terms how can we combine to just one and then it's perfect square
have a look we know that 1 minus sin Square Theta this right here gives us cosine Square
Theta because that's just a pag identity we know sin square plus cosine square is equal to 1 and
we just rearrange it and from here we can also derive the other ones for example we can also
say if we have secant Square Theta minus 1 this actually gives us tangent Square Theta and from
here we can also derive another one for example if I move the one to the other side we can say
1 + tangent Square Theta will give us secant Square Theta and these are the three main ones
as you can see we have 1 + something squared 1 + something squar and here we only end up with
one term squared aha that's the idea huh so I'm just going to start by say let X be tangent Theta
and then do the usual business here we have the DX so go ahead differentiate both sides
DX equals what's the derivative tent Theta is it there yeah secant Square Theta D Theta now
let's put this back here we will get the integral square root 1+ DX is tangent and then we have the
squared and then the DX is that so we have secant squar Theta D Theta so they multiplying like this
and now if you look at this this is really just secant squared so we have that and then of course
we still have the square root and this and that cancel so we just get see Theta but wait don't we
need the absolute value yes right but it's okay let me just tell you we don't need to here just
don't worry about it I know that's bad teaching but whatever anyway this is just secant Theta
times another secant Theta we are really just looking at the integral of secant thir power
Theta D Theta and if you did the homework already you will know the answer for this right here is 12
secant Theta * tangent Theta plus 12 Ln absolute value secant theta plus tangent Theta now we are
going to go back to the X world and to do so look at here we know tangent theta equals x which is X
over one go ahead and draw the right triangle here is the right angle and here's the angle
Theta and then for tangent we know that's opposite over adjacent so we have X over one like this and
for the hypotenuse is you first open the square root and you do this squared plus the other side
squared so from here we know what secant Theta is so now let's go back here we have 1/2 secant is
hypotenuse over adjacent so it's just Square < TK of 1 + x^2 and then for tangent Theta it's just X
over one just X and then let's just write down the rest 12 Ln absolute value actually we don't
need the Absol value because you will see secant is square < TK of 1 + X2 and then we add tangent
is X this is always positive so we just need a parenthesis done day five today I will show you
guys how to do partial fraction decomposition to break down this into smaller fractions that we can
actually integrate have a look let me write down the top right here 2x + 1 over for the
bottom we can actually Factor it right so let's go and Factor it we get x - 2 * x -1 now we can
break this down into two smaller fractions the first one will have x - 2 on the bottom and for
the second one we will have the xus one on the bottom notice both of them are linear meaning
we have X to first degree so that means the top will give us no degree so degree zero it's just
a constant constant a and constant B why because if you have the same degree let's say if you say
ax plus b then you can do long division you can break it down right so it's just constants all
right how do we find a let's use the so-called kafra method let me show you a is going to be
easy you'll see on the bottom we have x -2 we go back to the original and cover it up cover up the
X-2 and you ask yourself x minus 2 how can it be zero X has to be two right yeah so we put X to
B2 and we plug that into here and here not here because we covered it already so on the top we
have 2 * 2 and then + one over this two here and then minus one on the top I believe we get five
on the bottom I believe we get one and I believe 5 ided by 1 is equal to 5 so a is equal to 5 now
don't worry let's practice with b b we can also do cover up we have x - one come here cover it up
X is equal to 1 and we plug into here and here so we have 2 * 1 + 1 over 1 - 2 work this out we get
be3 so now we can come here and then just write down we have the integral this fraction is just
this right and now five over this we get linear right this is linear so it's natural log so we
have five times the natural log absolute value yeah and then the inside is x - 2 and then minus
3 * same thing natural log absolute value x - one plus C and that's it that's the answer yeah hey
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