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Say I have the indefinite integral $\int g(x)g'(x)~dx$, how would I go about solving this?

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2 Answers 2

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Notice that $2g'(x) g(x) = [g(x)^2]'$

So

$$\int g'(x) g(x) dx = \int [\frac{g(x)^2}{2}]' dx = \frac{g(x)^2}{2}$$

As the primitive of the derivative of a function is this function

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If we write $u=g(x)$ then $du=g'(x)dx$ and so

$$\int g(x)g'(x)~dx=\int udu=\frac{u^2}{2}+c=\frac{[g(x)]^2}{2}+c$$

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