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# 𝘶-substitution: defining 𝘶

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
A common challenge when performing 𝘶-substitution is to realize which part should be our 𝘶.

## Want to join the conversation?

• By the way, what's the antiderivative? I think it is [2(x²+x)sqrt(x²+x)] / 3
• 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
• -d/dx * (u*du/dx)+f=0
0<x<1
(Dy/dx) atx=0
U(1) =2^0.5
Can you help me to solve it
• 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.
• 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?
(1 vote)
• 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.
• 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

u/(1+u)=(u+1-1)/(u+1)=(u+1)/(u+1) -1/(u+1)=1-1/(u+1)
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.