Integration with respect to a higher power

[steven10137]

Homework Statement

<br /> \int {x^2 ,d(x^4)} <br />

Homework Equations

as a starter; the previous problem was:
<br /> \int {x^2 ,d(x^2)} <br />

and I managed to solve this by letting u=x^2 then integrating u:

<br /> \int {u} <br />

=<br /> \frac{x^4}{2}+C<br />

The Attempt at a Solution

can someone please explain the theory behind this?
my textbook gives no explanation and i don't really know what I am looking for...
I tried following the working from the previous problem through, but got (x^6)/3 when the answer was supposed to be 2(x^6)/3

thanks in advance
Steven

 

[Moridin]

In the previous problem, you made the differential to du. That is an interesting tactic for this problem too.[/color].

 

[steven10137]

so I am on the right track i take it?
If so;
\int {x^2 d(x^4)} = \int {u^2 d(u)}
am I correct?

 

[Moridin]

Not quite. If

u = x^4

then

x^2 = ?

 

[MathematicalPhysicist]

dx^4/dx=4x^3
dx^4=4x^3dx

 

[HallsofIvy]

In general,
\int f(x)dg(x)= \int f(x) \frac{dg}{dx} dx

 

[steven10137]

thankyou all for your help, I understand it now :)
dx^4 = 4x^3 dx
therefore
\int {x^2 dx^4}
= \int {x^2 4x^3 dx}
= 4 \int {x^2 x^3 dx}
= 4 \int {x^5 dx}
= 4 \frac {x^6}{6} + C
= \frac {2x^6}{3} + C

cheers
Steven