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I have been playing with the substitution rule in order to test some ideas with computational graphs. One of the things I'm doing is applying the substitution to well known, and easy, integrals. For example, let's use that method to find the indefinite integral for

$$f(x) = x^2$$

Using the rule $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, we get

$$F(x) = \frac{x^3}{3} + C$$

So, let's do the following substitution:

$$u = x^2$$

$$\frac{du}{dx} = 2x \Leftrightarrow dx = \frac{1}{2x} du$$

So, performing the substitution in the integral of $f(x)$ gives us

$$\int x^2 dx = \int u \frac{1}{2x} du = \frac{1}{2x} \frac{u^2}{2} + C = \frac{(x^2)^2}{4x} + C = \frac{x^3}{4} + C$$

Have I done anything wrong with the substitutions??

Thanks in advance!

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    $\begingroup$ Yes. You treated the function $\frac{1}{2x}=\frac{1}{2\sqrt u}$ as a constant. $\endgroup$ – user228113 Mar 20 '17 at 17:37
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$\int \frac{1}{2x}u du$ = $\int \frac{1}{2\sqrt{u}}\cdot u du$ = $ \int \frac{\sqrt{u}}{2}du$ = $u^{\frac{3}{2}}\cdot\frac{2}{3}\cdot \frac{1}{2} + C$

$= \frac{u^{\frac 3 2}}{3} + C = \frac{x^3}{3} + C$ as required.

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You have one mistake. If you are changing $dx$ into $du$, then you need to convert all terms containing $x$ into $u$.

So we have,

$$\int x^2 dx = \int u \frac{1}{2\sqrt u} du$$

$$= \frac 12 \int \frac{1}{\sqrt{u}}\cdot u du = \frac 12 \int \sqrt u du$$

$$= \frac{1}{2} \cdot u^{\frac{3}{2}}\cdot\frac{2}{3}+ C$$

$$= \frac{u^{\frac 3 2}}{3} + C = \frac{x^3}{3} + C$$

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if you set $$u=x^2$$ then we have $$x=\pm \sqrt{u}$$ and $$dx=\pm\frac{1}{2\sqrt{u}}du$$

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