Anti differentiation of f(x)f'(x)

[schlynn]

Homework Statement

$$\int$$(x³+x)[tex]\sqrt{x⁴+2x²}[/tex]dx

Homework Equations

$$\int$$[f(x)^r]f'(x)dx=(([f(x)]^r+1)/(r+1))+C

The Attempt at a Solution

The x³+x part is close to the derivative of the other part that I'm supposed to be anti differentiating, so should I just introduce a 4/4 into the equation and bring the 4 out front and then put the x³+x over the 4? That way it is the derivative? Then I can just say that the anti derivative is just 4[((x⁴+2x²)^(3/2)/(3/2))=8((x⁴+2x²)^(3/2)/(3))?
When I differentiate that though, I'm off by a factor of 16. I get an extra 16 out front.

Apparently it doesn't like nested latex?

 

[Mark44]

Fixed your LaTeX. Tip: don't put [ sup] tags inside [ tex] tags. For superscripts inside LaTeX code, use ^{}. Also, it looks better if you put the entire expression inside one pair of tex tags.

Homework Statement

$$\int(x^3+x)\sqrt{x^4+2x^2}dx$$

Homework Equations

$$\int$$[f(x)^r]f'(x)dx=(([f(x)]^r+1)/(r+1))+C

The Attempt at a Solution

The x³+x part is close to the derivative of the other part that I'm supposed to be anti differentiating, so should I just introduce a 4/4 into the equation and bring the 4 out front and then put the x³+x over the 4? That way it is the derivative? Then I can just say that the anti derivative is just 4[((x⁴+2x²)^(3/2)/(3/2))=8((x⁴+2x²)^(3/2)/(3))?
When I differentiate that though, I'm off by a factor of 16. I get an extra 16 out front.

Apparently it doesn't like nested latex?

You're on the right track.
Let u = x⁴ + 2x², so du = (4x³ + 4x)dx

With this substitution, your integral becomes
$$(1/4)\int u^{1/2} du = (1/4) (2/3) u^{3/2} + C = (1/6) (x^4 + 2x^2)^{3/2} + C$$

 

[schlynn]

Hmmmm, can you break it down into steps? I think I understand what you're saying, but the du in the problem, that's just 4x³+4x, and doesn't mean with respect to the variable u? It just looks like you multiplied through by 1/4, and not 4/4, don't you need to do 4/4 that way it doesn't change the equation?

 

[Mark44]

du = (4x³ + 4x) dx

In your problem you have (x³ + x)dx, and you need (4x³ + 4x)dx, so multiply (x³ + x)dx by 4 and also multiply the other part of the integrand by 1/4, which means you are multiplying the overall integrand by 4/4 = 1, which is always legal to do.

$$\int (x^3 + x)\sqrt{x^4 + 2x^2} dx = (1/4) \int \sqrt{x^4 + 2x^2} (4x^3 + 4x)dx$$
$$= (1/4)\int u^{1/2} du = (1/4) (2/3) u^{3/2} + C = (1/6) (x^4 + 2x^2)^{3/2} + C$$

 

[schlynn]

Ooooooohhhh, duh, don't know why I didn't see that before. I was basically swapping the 4 and the 1/4. Thanks a lot man.