# Integration with respect to a higher power

1. Jun 23, 2007

### steven10137

1. The problem statement, all variables and given/known data
$$\int {x^2 ,d(x^4)}$$

2. Relevant equations
as a starter; the previous problem was:
$$\int {x^2 ,d(x^2)}$$

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

$$\int {u}$$

=$$\frac{x^4}{2}+C$$

3. The attempt at a solution
can someone please explain the thoery behind this?
my textbook gives no explanation and i dont really know what im 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

Steven

2. Jun 23, 2007

### Moridin

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

3. Jun 23, 2007

### 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?

4. Jun 23, 2007

### Moridin

Not quite. If

$$u = x^4$$

then

$$x^2 = ?$$

5. Jun 23, 2007

### MathematicalPhysicist

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

6. Jun 23, 2007

### HallsofIvy

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

7. Jun 24, 2007

### 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