## Integrate wrt x^2

∫$\frac{1}{x^{2}+25}$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.
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 Recognitions: Homework Help 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.
 Recognitions: Gold Member Science Advisor 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.

## Integrate wrt x^2

I was wondering if we could carry out these operations with respect to curves aswell like the $x^{2}$.