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zeppBest ResponseYou've already chosen the best response.0
i've never seen a \(\large dx^2\), anyone can explain to me what it is?
 one year ago

saifoo.khanBest ResponseYou've already chosen the best response.0
Zippy! Me meet again. ;D
 one year ago

saifoo.khanBest ResponseYou've already chosen the best response.0
not sure how to solve this. :/
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
I recommend attempting the substitution: \(\sqrt{u} = x\). Just a hunch that it may turn out well (and not because I just explained the solution on irc chats). I think it may be helpful to our problem so that the f(x^2) dx^2 becomes f(u) du, etc...
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
\[ \newcommand\dd[1]{\,\mathrm d#1} \newcommand\de[1]{\frac{\mathrm d}{\mathrm d#1}} \text{Let } \sqrt{u} = x \text{. Then, } \\ \; \frac{1}{2 \sqrt{u}} \dd{u} = \dd{x} \\ \begin{align*} \implies \de{x} \int f(x^2) \dd{x^2} &= \frac{\text{d}}{\left(\frac{1}{2\sqrt{u}}\right) \; \dd{u}} \int f(u) \; \dd{u} \\ &= 2\sqrt{u} \de{u} \int f(u) \; \dd{u} \\ &= 2x f(u) \\ &= 2x f(x^2) \\ \end{align*} \]
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.1
Another Way: \(\de{x}\int f(x^{2})\;dx^{2} = \de{x}\int f(x^{2})\cdot 2x\;dx = \de{x} \left(F(x^{2}) + C\right) = f(x^{2})\cdot 2x\) Same thing, really.
 one year ago
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