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When integrating, say, $f(x) = x^{3} - 2x$, why do we go straight to integrating each term independently? Why is it not considered an implicit function of the power one and integrated as so?

I'm not sure if that's a valid feat since I'm a beginner, but I was thinking: $\int \left [ f(x) \right ] ^{^{1}} dx$ would be $\frac{f(x)^{2}}{2}$ as another example to the usual: $\int f(x)^{n} dx = \frac{f(x)^{n+1}}{n+1}$ where $n \neq -1 $.

As for what's inside, the $f(x)$, as I said I'm a beginner, but I know of u-substitution and maybe we could use it and work it out from there?

Is that possible or just plain wrong?

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The power rule for integration says that $$ \int x^n dx = \frac{x^{n+1}}{n+1} $$ Why doesn't that work for $\int [f(x)]^n dx$? Because $f(x)$ does not match $dx$. The proper use of the power rule would be $$ \int f(x)^n df(x) = \frac{f(x)^{n+1}}{n+1} $$ But we can still get something useful out of this by using, as you put it, some sort of chain rule. Namely, the chain rule, which tells us $df(x) = f'(x) dx$, which makes the above into $$ \int f(x)^n f'(x)dx = \frac{f(x)^{n+1}}{n+1} $$ As you may have noticed, this is just a $u$-substitution with $u = f(x)$. In fact, $u$-substitution is in general equivalent to the chain rule, and thus is the sort of chain rule you describe in the question.

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  • $\begingroup$ your answer was to the point but the correct integral of the last example is f(x) raised to the power of n+1 which you can check by differantiation . $\endgroup$ – sarah Mar 1 '17 at 21:30
  • $\begingroup$ Yes, that it is. $\endgroup$ – eyeballfrog Mar 1 '17 at 21:34
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You forget the chain rule. Indeed,

$$\frac d{dx}\frac{y^2}2=y\frac{dy}{dx}\ne y$$

Which is why integration is so much harder than differentiation.

However, it is true that

$$\int[f(x)]^nf'(x)\ dx=\frac{[f(x)]^{n+1}}{n+1}+c$$

Which should follow from u-substitution or differentiating both sides.

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It is wrong. The derivative of $[f(x)]^2/2$ requires you to use the chain rule, so $$\frac{d}{dx}\frac{[f(x)]^2}{2}=\frac{2[f(x)]^1}{2}f'(x)=f(x)f'(x).$$

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As the other answers have indicated, it is not sufficient to write a function in the form $[g(x)]^n$ in order to integrate it using the "power rule", even for $n=1$. One must express the function in the form $[g(x)]^n g'(x)$.

Luckily, your example function can be written in that form, for $n=1$! If $f(x) = x^3-2x$, we can cleverly choose $g(x) = \frac{\sqrt{2}}{2}(x^2-2)$, so that $g'(x)=\sqrt{2}\;x$ and $f(x)=[g(x)]^1 g'(x) = x^3 - 2x$. It follows that

$$ \int f(x)\,dx = \int [g(x)]^1 g'(x)\,dx = \frac{1}{2}\left[ g(x)\right]^2 + C = \frac{1}{4}(x^2-2)^2 + C = \frac{1}{4}x^4 - x^2 + 1 + C. $$

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