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When you have to integrate a function that requires substitution and you integrate it again, why is it wrong to keep the initial substitution?

e.g. $$y''=\frac{2x}{(1+x^2)^2}$$

If you let $u=1+x^2$ then $y'=-(1/u)+C$. Why is it wrong to integrate that again with respect to $u$ and then change back to $x$ at the end? I know it's not right but I can't see why

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Try your method and see if your final answer has second derivative as required. – Will Jagy Sep 19 '12 at 1:32
up vote 0 down vote accepted

Because of the Jacobian (or how is it called in 1 dimension?), I mean the factor between du and dx.

You want to integrate with respect to x, but when you keep the substitution, what you will do is integrate with respect to u.

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You want $\int y'~dx$, which is $$\int\left(\frac1u+C\right)dx\;.$$ This is $$\int\left(\frac1{1+x^2}+C\right)dx\;;$$ if you forget that you’re supposed to be integrating with respect to $x$ and simply write down $\ln|u|+Cu+D$ and convert back to $x$, you’re calculating $$\int\left(\frac1u+C\right)du\;,$$ which is $$\int\left(\frac1u+C\right)2x~dx=\int\left(\frac1{1+x^2}+C\right)2x~dx\;,$$ which you can see is a very different animal.

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You have $$\dfrac{d}{dx} \left( \dfrac{dy}{dx} \right) = \dfrac{2x}{(1+x^2)^2}$$ If you set $u = 1+x^2$, you get $$\dfrac{d}{dx} \left( \dfrac{dy}{dx} \right) = \dfrac{2x}{u^2} = \dfrac1{u^2}\dfrac{du}{dx} \implies d \left( \dfrac{dy}{dx} \right) = \dfrac{du}{u^2} \implies \dfrac{dy}{dx} = - \dfrac1u + C$$ Note that the left hand side is the derivative of $y$ with respect to $x$ i.e. $\dfrac{dy}{dx}$ and not the derivative of $y$ with respect to $u$ i.e. $\dfrac{dy}{du}$.

Hence, you need to either express $\dfrac{dy}{dx}$ in terms of $u$ and $\dfrac{dy}{du}$ or express $u$ in terms of $x$ and proceed.

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