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

Question

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.

Answer 1

It is simply a matter of integration by parts.

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 $v(x)$: $$ \begin{align} &\displaystyle\int g(x) \cdot g'(x) = \frac{g(x)g(x)}{2} \\ =& \boxed{\displaystyle\int g(x) \cdot g'(x) = \frac{g(x)^2}{2}} \end{align} $$

Answer 2

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 $\int g(x)g'(x) dx = g(x)g(x) - \int g'(x)g(x) dx \rightarrow 2 \int g(x)g'(x) dx = g(x)g(x) + C \rightarrow \int g(x)g'(x) dx = \frac{(g(x))^2}{2} + C_1$