Agebraic example of U-Substitution?

[tomas]

Could someone please show me a very easy agebraic example of U-Substitution?

 

[suffian]

Solve the following indefinite integral:
∫(2x+1)³dx

There are actually two ways for us to solve this problem:
(1) Expand the integrand and the integrate the polynomial.
(2) Perform a u-subtitution.

The easier method is clearly the second and it's what you want to know how to do anyway.

We find a subexpression to set u equal to (in this case it is fairly obvious):
u = (2x+1)

Then we find the derivative of this subexpression:
du/dx = 2

We multiply and divide the integrand by this derivative as follows (note that this simply means multiplying by one):
∫(2x+1)³ * (du/dx)/(2) dx
∫(1/2)(u)³(du/dx)dx

If you remember the chain rule for derivatives, then you'll also remember that the derivative of f(u(x)) = f'(u(x))u'(x). Well, Our integrand is already in this form now. To find its antiderivative, we can ignore the u'(x) and integrate f'(u) (w/ respect to u) to find f(u) and then simply change the formula back to f(u(x)) to obtain the final antiderivative.

Following through with this we obtain:
∫(1/2)u³du
(1/8)u⁴
(1/8)*(2x+1)⁴

 

[Sting]

Well someone beat me to it but another example can't hurt.

Suppose you have:

[inte][2x(x²+ 1)]dx

You can solve this by u-substitution because if you let u = x2 + 1, then du = 2x.

[inte][u^1/2]du = [(2u^3/2)/3] + C

One thing about u substitution is that you should pick a u where du is in the integrand.

 

[StephenPrivitera]

change of variable

I like to treat "dx"s and "du"s like factors in the equation. It makes the problem seem a lot more algebra-y, and people like algebra. Some people throw du's and dx's on then end because they know they should, not because they think it means something.
Anyway, here's a fun example.

[inte] xdx/(bx+a)
let u = bx+a
then, du=bdx
since no "bdx" exists in the integrand, you have to solve the equation for something that does, ie, dx
(In my math class, the teacher taught us to rewrite the integrand to make it fit the du/dx equation. I feel that it's a lot clearer to rewrite the du/dx equation to make it fit the integrand.)
dx=du/b
replace dx with du/b and bx+a with u
[inte] x(du/b)/(u)
since u=bx+a
x=(u-a)/b
replace x with (u-a)/b
[inte] [(u-a)/b](du/b)/(u)
From there, just a little algebra and you get:
1/b²[inte] (1-a/u)du= 1/b²[u-alnu]+C

 

[xRAWRx6]

Just in case anyone else falls upon this thread, i'll post one more problem.
(inte)(3x+4)^8 dx
u=3x+4
du=3dx
du/3=dx
plug your dx back into your original equation
(inte)(u)^8 (du/3)
1/3(inte)(u)^8 du
(1/3)(1/9)(u)^9 +c
(1/27)(3x+4)^9 + c