# Integrate wrt x^2

1. Dec 1, 2011

### labinojha

∫$\frac{1}{x^{2}+25}$d$x^{2}$
They were substituting x^2 for y ($x^{2}$=y) and thus the answer would come to be log(y+25)
that is log($x^{2}$+25)

I don't think this is the case , i guess that we would be differentiating wrt a 2nd degree curve like a parabola in case of this problem .

Would you people point out whats the real thing.

2. Dec 1, 2011

### Mute

If the integral is really as you wrote it,

$$\int d(x^2) \frac{1}{x^2+25},$$

then you are free to set y = x^2, as in this case your differential is d(x^2), and it already contains the x^2 so you are free to make the substitution.

If the integral were

$$\int dx \frac{1}{x^2+25},$$

then you can of course still substitute $y = x^2$, but then the differential term is different: $dx \rightarrow dy/(2x) = dy/(\pm 2\sqrt{y})$. Note that if this were a definite integral were the limits went from some negative value to a positive one, you would have to split the integral into two pieces, one from the negative value to 0 and one from 0 to the positive value, and then make the substitution, as you need to choose a sign for the square root and it's different for x > 0 and x < 0.

3. Dec 2, 2011

### HallsofIvy

Staff Emeritus
In general, if g(x) is a differentiable function, then dg(x)= g'(x)dx.

d(x^2)= 2xdx so
$$\int\frac{1}{x^2+ 25}d(x^2)= \int\frac{2x}{x^2+ 25}dx$$

Now, if you let $u= x^2+ 25$, du= 2xdx and the integral becomes
$$\int\frac{du}{u}= ln(u)+ C= ln(x^2+ 25)+ C$$
just as before.

That really is the case.

4. Dec 2, 2011

### labinojha

These integrations and differentiations are done in with respect to the x-coordinate which is a straight line .
I was wondering if we could carry out these operations with respect to curves aswell like the $x^{2}$.

Sorry if this is absurd, but i would be glad if i was clarified .

5. Dec 3, 2011

### HallsofIvy

Staff Emeritus
Yes, you integrate a function f(x) with respect to another function g(x). You are, in effect, "changing the scale". For the Riemann integral, such a function, g, would have to be differentiable and, as I said before, [itex]\int f(x)dg(x)= \int f(x)g'(x)dx[/tex]. For the Stieljes integral, g does not have to differentiable, only increasing.