AP®︎/College Calculus BC
- 𝘶-substitution intro
- 𝘶-substitution: multiplying by a constant
- 𝘶-substitution: defining 𝘶
- 𝘶-substitution: defining 𝘶 (more examples)
- 𝘶-substitution: defining 𝘶
- 𝘶-substitution: rational function
- 𝘶-substitution: logarithmic function
- 𝘶-substitution warmup
- 𝘶-substitution: indefinite integrals
- 𝘶-substitution: definite integrals
- 𝘶-substitution with definite integrals
- 𝘶-substitution: definite integrals
- 𝘶-substitution: definite integral of exponential function
𝘶-substitution: defining 𝘶
A common challenge when performing 𝘶-substitution is to realize which part should be our 𝘶.
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- By the way, what's the antiderivative? I think it is [2(x²+x)sqrt(x²+x)] / 3(4 votes)
- I got (2/3)[(x² +x)^(3/2)]+c I think that when you took the square root of "u", you accidentally dropped the exponent of 3. And of course, don't forget to include c!
In your format I think it would look like 2(x²+x)sqrt[(x²+x)^3] / 3 + c(13 votes)
- This is all nice and tidy...but what if you don't have that convenient 2x+1 term in the front? Sal makes it a point to say that you need to have a matching derivative for you to do the U Subbing...so what if this wasn't there?(2 votes)
- Then you would need to find a different integration technique. There are a few other cases you'll see on Khan Academy like integration by parts and trigonometric substitution.
But you should remember that there are some integrals, like e^(-x^2), that simply cannot be computed except by approximation.(3 votes)
- What is the final answer to this problem? Is it, [2/3(x^2+x)^3/2](2x+1)+C ?(1 vote)
- It is not, it is 2/3 * sqrt[(x²+x)³] + C Look at the comments below.(4 votes)
- -d/dx * (u*du/dx)+f=0
Can you help me to solve it(2 votes)
- These examples work when there already is u'(x)*f(u(x)), or when u-substitution is easily made when all there is missing is a multiplied scalar.
How would you operate the indefinite integral of 1/(1+sqrt(x))?
If you make u equal to 1+sqrt(x), du = 1/(2*sqrt(x))dx which doesn't appear in the equation at first. And if I try to add du inside the integral and compensate it by (2*sqrt(x)). Finally, I would simplify it which would give me
the integral of 2*sqrt(x)*1/u du...which seems wrong, since I should only be having one variable. How would u substitution apply here??(1 vote)
- Substitute u=√x, so du= 1/(2√x) dx=1/(2u) dx. Rearranging, dx=2u·du
Now ∫(1/(1+√x))dx=∫(1/(1+u)) ·2u·du=2∫(u/(1+u)) du
Now you can solve the integral by splitting it across the two terms and back-substituting.(1 vote)
- So would substitution only be used when you have a term that is the derivative of the other?(1 vote)
- Pretty much however substitution can be used in many case even if you can`t apply something like reverse chain rule such as in the case of (1-x^2)^1/2 you could substitute
u= sin x alternatively you may make t-formula substitution so you bring an expression to some algebraic form so you could split it up using partial fraction. There is also integration parts although in that case you would substitute u= G(x) so you can integrate f(x)g(x) using a formula similar to the product rule.(1 vote)
- How are we supposed to know when to apply the Chain Rule in reverse as opposed to the Product Rule or Quotient Rule in reverse? sin(x)/cos(x) looks like a fraction — it doesn't pattern match against f'(g(x))g'(x).
I tried "pausing the video" and attempted to solve the second problem but didn't succeed. When I saw Sal's explanation, I didn't feel like I learned a general lesson about how to solve integrals — instead, I learned some stupid one-off trick about tangent and a lesson about the futility of applying general principles.(0 votes)
- [Tutor] What we're going to do in this video is give ourselves some practice in the first step of u substitution, which is often the most difficult for those who are first learning it and that's recognizing when u substitution is appropriate and then defining an appropriate u. So let's just start with an example here, so let's say we wanna take the indefinite integral of two x plus one times the square root of x squared plus x, dx, does u substitution apply here and if it does, how would you define the u? Pause the video and try to think about that. Well, we just have to remind ourselves, that u substitution is really trying to undo the chain rule, if we remind ourselves what the chain rule tells us, it says look, if we have a composite function, let's say f of g of x, f of g of x and we take the derivative of that with respect to x, that that is going to be equal to the derivative of the outside function with respect to the inside function, so f prime of g of x times the derivative of the inside function, times the derivative of the inside function and so u substitution is all about well, do we see a pattern like that inside the integral? Do we see a potential inside function, a g of x, where I see its derivative being multiplied? Well, we see that over here, if I look at x squared plus x, if I make that the u, what's the derivative of that? Well, the derivative of x squared plus x is two x plus one, so we should make that substitution, if we say u is equal to x squared plus x, then we could say du dx, the derivative of u with respect to x is equal to two x plus one, if we treat our differentials like variables or numbers, we can multiply both sides by dx, which is a little bit of hand wavy mathematics, but it's appropriate here, so we could say two x plus one times dx and now what's really interesting here is we have our u right over there and notice we have our two x plus one times dx, in fact it's not conventional to see an integral rewritten the way I'm about to write it, but I will, I could rewrite this integral, you should really view this is the product of three things, oftentimes people will just view the dx as somehow part of the integral operator, but you could rearrange it, this would actually be legitimate, you could say the integral of the square root of x squared plus x times two x plus one dx and if you wanted to be really clear, you could even put all of those things in parentheses or something like that and so here, this is our u and this right over here is our du and so we could rewrite this as being equal to the integral of the square root of u, 'cause x squared plus x is u, times du, which is much easier to evaluate. If you are still confused there, you might recognize it if I rewrite this as u to the one half power, because now we could just use the reverse power rule to evaluate this and then we would have to undo the substitution, once we figure out what this anti derivative is, we then would then reverse substitute the x expression back in for the u.