There's a whole family of solutions (x^2 + 9)^n/2 where n is any positive odd integer. The integrand then evaluates to (x^2 + 9)^m where m = (n-1)/2 is an integer.
Just come up with whatever problem you want and multiply it by sqrt(x^2+9). You've replaced the problem with an unrelated problem, and so long as your new problem doesn't require a trig substitution, you're golden.
Problems like this prove to me how mathematically basic I am. A decent mathematician but not a great one. Because I immediately thought of (1) as a solution, but didn’t even think of the hilariously simpler (2)…
1 is also okay since you can use hyperbolic substitution, which is not a trig sub also, there is Euler substitution to deal with it so trig sub is really not necessary
Technically, there are infinitely many answers to this problem. First and most obvious, you could fill in the blank with any K(sqrt(x^2+9)) such that K is a constant, so that the roots would cancel each other out and the resulting integral would end up being equal to Kx+C Secondly, it is possible to fill it in with any f(x)(sqrt(x^2+9)) and then only calculate the integral fo f(x)dx, as long as it by itself doesn't require a trig substitution.
Another infinite set of solutions is just to have any integrable function which doesn't require trig substitution *a copy of the denominator, e.g. e^(2x)*sqrt(x^2+9) as the blank
what about the denominator being to the -1/2 power and then doing a numerator with the same base and combining powers? can get the same answer of 1 or bring it to any power that you want and then integrate. its still not a trig sub.
The general quadratic solution I found: (a*x^2 + b*x + c), where 2*c = 9*a If you carry it out for this general form, you'll see a solution of: 1/2*sqrt(x^2 + 9)*(a*x + 2*b) + (c - 9/2*a)*asinh(x/3) + K Any solution where 2*c = 9*a, will eliminate the asinh term. You can then assume (p*x + q)*sqrt(x^2 + 9) is the solution, and differentiate the ansatz to match coefficients.
What is the issue? th-cam.com/video/3RDMnFLeE_g/w-d-xo.html
Blank being sqrt(x^2+9) it's a 300 IQ move.
Well, the question didn't say we can't just put "zero" into the box!
WRONG 🤣
Or sqrt(x^2+9)
I would just put the derivative of the denominator in the numerator, so it becomes integral of f'/f which is just ln(f)+c
My first instinct, too. Easy.
Your expression should be ln|f| + c though. In the ex in the video, doesn't matter bc sqrt(x² +9) is positive for all R
another solution could be (x²+9)^3/2. The expression simplifies to just x²+9, which becomes x³/3+9x+c, no trig substitution in sight.
There's a whole family of solutions (x^2 + 9)^n/2 where n is any positive odd integer. The integrand then evaluates to (x^2 + 9)^m where m = (n-1)/2 is an integer.
Any power of the denominator could work in that case. Even integers. 2n+1 / 2 is convenient because you get an integer power upon simplification
Just come up with whatever problem you want and multiply it by sqrt(x^2+9). You've replaced the problem with an unrelated problem, and so long as your new problem doesn't require a trig substitution, you're golden.
sqrt(x^2 + 9) is the immediate solution I came up with as well :D
I love how he looks at the camera angrily knowing that the students who came up with those solitions are watching
Problems like this prove to me how mathematically basic I am. A decent mathematician but not a great one. Because I immediately thought of (1) as a solution, but didn’t even think of the hilariously simpler (2)…
1 is also okay since you can use hyperbolic substitution, which is not a trig sub
also, there is Euler substitution to deal with it so trig sub is really not necessary
Technically, there are infinitely many answers to this problem.
First and most obvious, you could fill in the blank with any K(sqrt(x^2+9)) such that K is a constant, so that the roots would cancel each other out and the resulting integral would end up being equal to Kx+C
Secondly, it is possible to fill it in with any f(x)(sqrt(x^2+9)) and then only calculate the integral fo f(x)dx, as long as it by itself doesn't require a trig substitution.
Another infinite set of solutions is just to have any integrable function which doesn't require trig substitution *a copy of the denominator, e.g. e^(2x)*sqrt(x^2+9) as the blank
Doesn't say it has to have been simplified, so 0 or f(x)*(x^2+9).
How about finding the set of all possible answers.
It contains infinite elements
f(x) × sqrt(x^2+9) where f(x) is any function that doesn't need a trig sub to integrate
What about putting the same thing on the top? Integrate 1
The second one was what I was thinking at first. Just divide by the same thing to make integral of 1.
what about the denominator being to the -1/2 power and then doing a numerator with the same base and combining powers? can get the same answer of 1 or bring it to any power that you want and then integrate. its still not a trig sub.
The general quadratic solution I found:
(a*x^2 + b*x + c), where 2*c = 9*a
If you carry it out for this general form, you'll see a solution of:
1/2*sqrt(x^2 + 9)*(a*x + 2*b) + (c - 9/2*a)*asinh(x/3) + K
Any solution where 2*c = 9*a, will eliminate the asinh term. You can then assume (p*x + q)*sqrt(x^2 + 9) is the solution, and differentiate the ansatz to match coefficients.
another answer would be to let the numerator equal to any constant multiple multipled by x raised to an odd number
Please clarify what is ♾️ power i
Not defined!
Last 2 answer were amazingly crazy
use y=x^n where n€ positive odd integers
Just put sqrt(x²+9) times anything that on its own would not need trig substitution
Answer: 2sqrt(x²+9) + c
We can put sin(x) or e^x then the function becomes non integrable
I immediately thought of finding the derivative so we can integrate to ln(sqrt(x^2+9)) lol
Couldnt it be x•(x²+9)^(-½) leaves you with just ln(√(x²+9)) + C or ln(x²+9)/2 + C
I think this question is to remind students not to overthink. ‘1’ or ‘0’ were probably the intended answers.
I would put blank = sqrt(x^2+9) * 2x * f'( x^2+9) for any function f to get the answer f(x^2+9) for any function f.
The blank could be f'(x)(x²+9)½ so the integral is f(x), f(x) is differentiable function. For example, f(x) = constant, polynomial, ...
The Werefrog immediately thought of the second answer, after a bit of thought, the 0 also came up. Didn't think of the u substitution, though.
x^2 +9 in the numerator?
That would also require trig sub
@epikherolol8189 you could use the power rule, though as it simplifies to integral of sqrt(x^2+9)dx.
@@hipeplefulu will still need trig sub from there
@@JashXD why? I thought trig sub was only for when the denominator has a square root?
@@hipepleful how would you integrate √(x²+9) otherwisr
Enter 1 and recognise it as an elementary integral.
I havent started integration yet but these look fun!
Did you understand the video?
The derivative of sqrt(x^2+9) can be put so we end with ln(x^2+9)+C
Put sqrt(x^2 + 9). Then you are just intergrating dx 😁.
Whoops I made this comment before watching the video 😂
0
if you put x(x²+9)^(-1/2) in the box, then the answer is just 1/2*ln|x²+9|+C
2x will also work
And 2x/√x2+9 too
Being this early feels illegal
+1
It is. I've called the FBI and they're on their way.
put sqrt(x^2+9) in the numerator LOL
Why not just write sqrt(x²+9)in the blank🤔
I would just write the numerator as it is lol
My first guess would be sqrt(x^2 + 9) 😂 then we have the integral of 1, no u sub needed
kx*e^(sqrt(x^2+9)), k some real number.
just put x in the black and than the answer is sqrt(x^2 + 9) automathicly
What happen to your poke ball
I will just put 0 coz why not😂
Maybe lazy but just put zero
Don't call that "lazy", that's being smart
It does check all the requirements
2√u +c
=2√(x²+9)+ c
dy