# Indefinite integral of g(x) * g'(x)

I'm given the question
$$\int g(x) * g'(x) dx$$
The integral evaluates to
$$1/2[g(x)]^2 + C$$
but I have absolutely no idea how to arrive at that conclusion.

My first thought was that the $$g(x)*g'(x)$$ looked somewhat similar to the product rule, and that I could somehow use that with the fundamental theorem of calculus to arrive at the answer, but I had no luck. I see multiple questions like this in my calc textbook and I never really know how to approach them.

• A simple substitution approach... $y=g(x),dy=g'(x)dx$, then you have $\int ydy$... Sep 17 '21 at 3:17
• Just observe that $\frac12\bigl(g^2(x)\bigr)'=g(x)g'(x)$. Sep 17 '21 at 16:44

Our integral is as such: $$\displaystyle\int g(x) \cdot g'(x)$$ We can rename this, and then use integration by parts, like so ($$u(x)$$ and $$v(x)$$ both represent $$g(x)$$): \begin{align} & \displaystyle\int g(x) \cdot g'(x) \\ = & \displaystyle\int u(x) \cdot v'(x) \\ = & u(x)v(x) - \displaystyle\int u'(x) \cdot v(x) \end{align} Now rearranging: $$\displaystyle\int g(x) \cdot g'(x) = u(x)v(x) - \displaystyle\int u'(x) \cdot v(x) \\ 2\cdot\displaystyle\int g(x) \cdot g'(x) = u(x)v(x) \\ \displaystyle\int g(x) \cdot g'(x) = \frac{u(x)v(x)}{2}$$ Substituting in for $$u(x)$$ and $$v(x)$$: \begin{align} &\displaystyle\int g(x) \cdot g'(x) = \frac{g(x)g(x)}{2} \\ =& \boxed{\displaystyle\int g(x) \cdot g'(x) = \frac{g(x)^2}{2}} \end{align}
• Note that you can also do this with a $u$-substitution, but this is the first approach that came to my mind Sep 17 '21 at 3:25
You can use the product rule: $$\int uv'dx = uv - \int u'vdx$$, so we let $$u = g(x), v' = g'(x)$$ then $$u' = g'(x), v = g(x)$$
Then $$\int g(x)g'(x) dx = g(x)g(x) - \int g'(x)g(x) dx \rightarrow 2 \int g(x)g'(x) dx = g(x)g(x) + C \rightarrow \int g(x)g'(x) dx = \frac{(g(x))^2}{2} + C_1$$