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$$\int (f(x))' dx = f(x) + c$$

if $u=g(x)$ then

$$\int (f(u))'du = f(u)+c$$


$$\int (f(g(x)))'dx = f(g(x))+c$$

Where did I go wrong?

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You replaced $u = g(x), du = dx.$ If $u = g(x)$ then $du = g'(x) dx.$ – user2468 Aug 3 '12 at 4:10

In short, you forgot the chain-rule. That is to say that if $u = g(x)$, then $du = g'(x)dx$. It is this that let's us say that

$$ \int [f(g(x))]'dx =\int f'(g(x))g'(x)dx = f(g(x)) + C$$

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