in 2nd step , let x³+1=t , dx =dt/(3x²) and x³=(t-1) putting all these value in 2nd line and solving equation will be ⅓(int. (t^5/2 -t½) dt) and it can be easily solve in 3 lines 😎😀😀
Dear Jaden, we are doing integration by parts. Therefore, we have to be very careful when splitting the factors. My students in the classroom tried to use x^4 and they got stuck. There is no set rule for picking the right exponents, it comes with lots of practice and drill. Hope I answered your question to your satisfaction. I'm sure you are an awesome and smart student 👍 Please watch some more premath videos on integration by parts and you'd feel more comfortable with these kind of problems. Stay blessed and healthy😃 Have a very happy and blessed New Year!
You could do that, but we have to be careful here about using derivatives that have things in common. We need to find x^5 and x^3 and see what derivatives have in common. We need to split x^5 as (x^3)(x^2) because we see here that the derivative of x^3 is 3x^2. The integral has to split apart as (x^3)(x^2)√(x^3+1). Putting in the u-sub (v-sub in this video here) for x^2√(x^3+1) by letting u=x^3+1 (i.e. v=x^3+1) will have du=(3x^2)dx (dv=(3x^2)dv) meaning that the integral being substituted to the u term will be (1/3)u^(3/2) (1/3)v^(3/2) and then evaluate that integral from there. If I split x^5=x(x^4), then u-sub doesn't work since there's nothing to put in terns of du and no cancellation happens.
Helpfull , thanks alot 👏👏👏
Thank you so much! Please keep supporting my channel. Kind regards 😀
i applaud the pace and explanations for everything. As a visual learner I have to see things pan out so this was great! Thanks
but why not use u = x^5 and dv = (x^3+1)^1/2 ?
THANKS PROFESOR !!!!, VERY INTERESTING!!!!!!!
Thank you so much, I was given the exact question
in 2nd step , let x³+1=t , dx =dt/(3x²) and x³=(t-1) putting all these value in 2nd line and solving equation will be ⅓(int. (t^5/2 -t½) dt) and it can be easily solve in 3 lines 😎😀😀
Ok. This is just the second type of binomial irrational integrals, and this procedure is imho better than integration by parts
Excellent video!!
This is really helpful!!
How do you know how to split the exponents. Why didn’t you use x^4 and x for example ...
Dear Jaden, we are doing integration by parts. Therefore, we have to be very careful when splitting the factors. My students in the classroom tried to use x^4 and they got stuck. There is no set rule for picking the right exponents, it comes with lots of practice and drill. Hope I answered your question to your satisfaction. I'm sure you are an awesome and smart student 👍 Please watch some more premath videos on integration by parts and you'd feel more comfortable with these kind of problems. Stay blessed and healthy😃
Have a very happy and blessed New Year!
You could do that, but we have to be careful here about using derivatives that have things in common. We need to find x^5 and x^3 and see what derivatives have in common. We need to split x^5 as (x^3)(x^2) because we see here that the derivative of x^3 is 3x^2. The integral has to split apart as (x^3)(x^2)√(x^3+1). Putting in the u-sub (v-sub in this video here) for x^2√(x^3+1) by letting u=x^3+1 (i.e. v=x^3+1) will have du=(3x^2)dx (dv=(3x^2)dv) meaning that the integral being substituted to the u term will be (1/3)u^(3/2) (1/3)v^(3/2) and then evaluate that integral from there. If I split x^5=x(x^4), then u-sub doesn't work since there's nothing to put in terns of du and no cancellation happens.
nice question
you should say can you integrate this function it is not a equation
thanks a lot!
A bit late to comment but i solved it with substituting u=sqrt(x^3+1)!