how to solve hard integrals!
it doesn't seem like we can just do an easy substitution
like if i let u equal to the inside it's just a no if i let you equal inside it's a no
i don't know but first i would like to tell you this question is from the mit integration b and
this question is from the book called putnam and beyond both of them are like great stuff of course
but you just have to think about all the possibilities
in order to solve all this kind of questions once you see it you'll see the solutions pretty
short but it's the first move that's the hard part which practice you can get better right
all right let's let's do this have a look inside we have x to the fourth power plus one huh so
let's do the following well suppose i have this expression let's say x to the fourth power i'm
not going to put on plus one but rather i'll put on plus x imagine i put this down what can
i possibly ask you i can ask you to factor this sure if we do that we just get x times x to the
third power plus one great and we usually factor out x to the smaller power like that okay but can
we also factor out x to the bigger power yes it's not usual but you know it's possible we can factor
x to the fourth power and becomes one plus you can do x divided by x to the fourth which is x to the
negative three perfect and that's exactly what we are going to do for this right here so check this
out i'm going to look at the inside right here i will factor out x to the bigger power which is
just x to the fourth times x sorry x to the fourth then factor that already so it's just one plus
1 divided by that which is x to the negative 4 like this okay now have a look this right here
we will have the integral and we have the 1 over we still have the x squared in the front
hmm we have this thing to the third power this thing to a third power that's kind of bizarre
but don't worry because what we can do is all right you see technically x to the first power
instead of x inside of the fourth root they cancel and then we just have raised this to the third
power so it's pretty good right so let's go ahead and put down x to the third power right again x
to the third power but this thing is still inside of that so we still have to write that down namely
the fourth root of parentheses 1 plus x to the negative 4 and then raised to the 3rd power here
if any of you guys did this with tricks up whatsoever let me know but anyway this is
what we have now check this out this times this is x to the 5th power i can actually
bring that up and this becomes the integral x to the negative fifth power over the fourth root of
the inside 1 plus x to the negative 4 and then raised to the third power like this
and this is actually so good because now we can just let the u equal to the inside have a look
let u equal 1 plus x to the negative 4 and you see d u is equal to bring the negative 4 to the front
minus 1 so it's negative 5. and we have negative 5 over there isn't it yes definitely and now
of course don't forget the dx right perfect now as usual let me isolate the dx which is
d u over negative 4 x to the negative 5 like that and i'll clean things up a little bit this is the
integral here we have x to the negative five over here i have u and then to the three over
four power pretty much right so i will write it as u and again this is the same as say three over
four power all right and then the dx is that which is d u over negative four x to the negative five
now the best thing happened which is the x minus five cancellation oh my goodness that's so great
so right here we see we have the negative 1 over 4 in the front bring this up this is u
to the negative 3 over 4 and then d u of course add 1 which is going to be positive 1 over 4
divided by 1 over 4 which is 4 over 1 and you see they cancel nicely so ladies and gentlemen
the answer to this integral is we have the negative and this is the fourth root of the
u which is that so i just put this down 1 plus x to the negative fourth power like this
all right now for this one do we do the same thing
notice earlier it works out really nicely because
the negative powers right negative 4 minus 1 is negative 5 so that's the reason huh
but it seems like we don't deal with any negative powers right here at all
well again a lot of this is that if you know the first move if you just try it and if it works
it works right the deal is that x 26 plus x to the third power well
if i borrow an x right here and put it inside i will be able to produce something inside that
actually we can do usa from there because if you want to u sub right now
the power inside is not big enough
the derivative of x plus power is just 3x squared right this is x to the sixth power that's no good
but have a look i'm going to write this as parentheses with x to the fifth plus x squared
and then i will multiply by x right because i want to just factor on x and then you see we still have
this x cubed plus 2 instead of the cube root dx and what i will do is i'm going to put this inside
and you see x to the first power outside of the cube root when you bring that inside it's
actually x to the third power third power times this hey it's x to the sixth power hey that works
similarly that will also work for this one so check this out right here we will end up with
the integral parentheses x to the fifth plus x squared and again this becomes the cube root
we'll bring this inside we get x to the sixth power and then plus 2 x to the
third power very very nice and then the dx all the way at the end right here of course now
right here let's just do a use up let u equal to x to the sixth plus two x to the third power and
you see d u is equal to the derivative of that is 6 x to the fifth haha and then this right
here is plus 6 x to the second aha and then of course we still have the dx right here right
so now you see this is just the integral of x to the fifth plus x squared and then
this is the cube root and i'm just going to write the cube root as the one third power
and then the dx is you just divide down both sides so you get d u over six x to the fifth plus six x
squared well you can factor out the six and x to the fifth plus x to the second and this and that
can solve very nicely so only o we just have to integrate well we still have the one over six
but we have to integrate this is u to the one third power and then we have the d u add 1 to
this which becomes 4 over 3 i believe yes and then divided by that which is 3 over 4
i believe yes and then that's going to be 1 over 8 yes and then this is man the cube root
of and then the u is what this guy which is x to the 6 plus 2 x to the third power and then
raised to the fourth power i'm so sorry about this raised to the fourth power
all right like this right so yeah and then yes the plus c yeah just like that
and i actually saw the solution for this otherwise i wouldn't be able to figure this out
but in fact i did this a few years ago already so i would just say this is like you know the
famous steps right not the famous one the famous first move keep these things in mind
because maybe later on will you encounter an integral maybe in your dream like somebody
when you are sleeping and then the integral is trying to attack you you know which measure to
use so you can you know have a good night's sleep and other stuff but anyway as always that's it