Good question

Excellent

2^11*ln2=ln2*(x-11)*e^((x-11)*ln2) , 2^3*2^8*ln2=ln2*(x-11)*e^((x-11)*ln2) , 8*ln2*e^(8*ln2)=ln2*(x-11)*e^((x-11)*ln2) , 8*ln2=ln2*(x-11) , 8=x-11 , x=3 , test , 2^3+3=11 , W( 2^11*ln2)=~ 5.545175444 , ln2*(11-x)=~ 5.545175444 , x= -(5.545175444/ln2)+11 , x=3 ,

2^x=11-x 1=(11-x)2^x 2^11•ln(2)=(11-x)ln(2)e^(11-x)ln(2) (11-x)ln(2)=ln(256) 11-x=log_2(256)=8 x=11-8=3

Excellent

Nice job. That was slightly headachy, but not too bad ; )

If you want integral to calculate try this \int \frac{x}{\sqrt{(x+2)^2+exp(x)}}dx Result should be x - 2ln(x+2+\sqrt{(x+2)^2+exp(x)})+C and it is quite interesting because programs like Mathematica or Wolfram alpha can't calculate it You can try this and record video about it

x=3 is a solution but is it the only one

After seeing this my brain 🧠: please pull out calculator 100ms

The x=3 solution is visible,after that you could move all the terms of your equation to one side and define a function.From there you could prove that the function is monotonous through its derivatives and thus x=3 is a unique solution

Wow. :0

I=L{sin(3x)/x}|[s=2] L{f(t)/t}=int[s,♾️](L{f(t)})ds I=int[2,♾️](L{sin(3x)})ds L{sin(ax)}=a/(s^2+a^2) I=int[2,♾️](3/(s^2+9))ds I=(arctan(s/3))|[2,♾️] I=pi/2-arctan(2/3) I=arctan(3/2)

Sir is there any method we can use Laplace in this

e^ ➖ 2x sin(3x)^2/(x)^2 =e^ ➖ 2x{sin*9x^2}/x^2=w^{In^s^n^x^dx^sin^s^n ➖9x^2/dx^s^n ➖ x^2In^s^n^x^dx^sin^s^n^ ➖ 9x^2/dx^s^n ➖ x^2}^w=w^{In^s^n^x^dx^sin^s^n ➖ 9x^4/dx^sins^n ➖ 9x^4In^s^n^x^dx^sin^s^n^dx^sin^s^n ➖ 9x^4/dx^sin^s^n ➖ 9x^4}^w=w^{In^s^n^x^dx^sin^s^n^ ➖ 3^2x^2^2/dx^sin^s^n ➖ 3^2x^2^2In^s^n^x^dx^sin^s^n ➖3^2x^2^2/dx^sin^s^n ➖ 3^2x^2^2}^w=w^{In^s^n^x^dx^sin^s^n ➖ 1^1x^1^1/dx^sin^s^n ➖ 1^1x1^1In^s^n^x^dx^sin^s^n^ ➖ 1^1x^1^1/dx^sin^s^n ➖ 3^1x1^2}^w=In^s^n^x^dx^sin^s^n ➖ 3x^2 (w^{In^s^n^x^dx^sin^s^n^➖ 3piIn^s^n^x^d^x^sin^s^n^x +2}^w).

Nice work ; )

I Feynmanned this one instead, with e^(- ax) then did the Euler form of sin(3x). One thing I missed was the neat use of arctan(1/x) at the end - I like that.

I would use integration by parts u = sin^2(x) , dv = 1/x^2dx then use double angle formula sin(2x) = 2sin(x)cos(x) then we can use substitution t = 2x Finally calculate L(sin(t)/t) and plug in s =0

Excellent use of the fundamental properties of Laplace Transform 🍻

Could you do Lobachevsky's rule on this?

cool integral, tried to solve it myself but changing variable mid integration always throws me off a bit

In^x^e^x(x)^2In^x^e^x/In^x^e^x{x+x ➖}+In^x^e^x{1+1 ➖}=x^2/In^x^e^x{x^2+2}w^{In^x^e^xdx^n ➖ 2/dx^2n ➖ 2In^x^e^xdx^ n ➖ 2/dx^2n ➖ 2}^w =w^{In^x^e^x^dx^2n ➖ 4/dx^2n ➖ 4In^x^e^x^dx^2n ➖ 4/dx^2n ➖ 4}^w=w^{In^x^e^x^dx^2n ➖ 2^2/dx^2n ➖ 2^2In^x^e^x^dx^2n ➖ 2^2/dx^2n ➖ 2^2}^w=w^{In^x^e^x^d^x^1n ➖ 1^1/dx^1 n ➖ 1^1In^x^e^x^dx^1n ➖ 1^1/dx^1n ➖ 1^2}^w=dx^1n ➖ 2 (w^{In^x^e^xdx^n ➖ 2piIn^x^e^x^dx^n+1}^w). i liked the video i watched enjoyed thank fir Laplace transformation method substituted.

I haven't studied laplace transform so I'm wondering can you also do it with integration by part ( talking abt ue^-un)?

We could expand 1/(1+x) into a geometric without the exponential substitution to get the result.

Excellent! Two of my favorite techniques - geometric sum and Laplace🍻

Nice, very similar to the one you did on owl3.

This is something 🎉

In my opinion the best way to calculate this integral is to factor out sqrt(5) from denominator of the integrand and use substitution sqrt(x^2+6x+8) = u - x But if we want result presented on video it would be better to use Euler substitution with roots of quadratic trinomial inside square roots

nice solution, thank you very much

Excellent

When the formula doesn’t work, does that necessarily mean the integral will diverge?

Crazy substitution 👏👏👏 loved the reflection formula🍻

Very interesting approach.

It is not hard to see that any integral (from 0 to inf.) with integrand a^[-x^p] * x^q ,with a >1, p > 0, q > -1 can be expressed with the Gamma function .

The most underrated math channel on yt

An elegant way to calculate this integral is to use Cauchy 's integral theorem, stating that the integral of f[z] over a closed path P in the complex plane vanishes if P does not enclose any singularities of f[z]. We write the integral as Re[1/2 ∫exp[-x^2]*exp[i x]]=1/2*Re [exp[-(x- i/2]]*exp[-1/4],and the integration is from -inf. to + inf. Take f[z] = exp[-z^2] and consider a rectangular path in C : P1 = {- N < z = x < n} ,P2 = {z = N+i*y ,(0< y < 1/2)}, P3 = {z = i/2 +x ,(N > x > -N)}, P4 = {z = - N +i *y ,( 1/2 > y > 0 ) . If we let N -> inf. ,then it easy to show that the integrals along P2 and P4 vanish. On P1 we have ∫exp[-x^2 ] =√π , on P3 we have - integral we want to calculate . Since f[z] does not enclose any singularities the sum of our remaining integrals is zero , this means that the integral = 1/2 *√π *exp[ -1/4]

Beautiful 🥰🥰

You ruin anything good in your video with your way of speaking.

Speak clearly. You mumble with your tongue and your hand.

As an engineer point of view, you could consider e=3 and proceed from there to have e^(-x^3) for an approximated result. Nice video.

Not satisfied with that solution

Great job! ; )

Excellent 🍻

Thanks for sharing!

Lobachvsky saved us.

Yes! Definitely Lobachevsky's 🍻

Huff pray to God hope i don't want to face infinity problems type in my exams

We can also use many alternatives for this

Fantastic

This integral is challenging without the lnx substitution and the cancellations.

Excellent