Find $\int g(x)g'(x)\text{ dx}$ where $g'(x)$ is continous

Question

$$\int g(x)g'(x)\text{ dx}$$

I'm not sure how to evaluate this.

I know $\int g(x)\text{ dx} = G(x)$, and I know $\int g'(x)\text{ dx} = g(x)$,

but how can I break up this integral to solve the main integral?

Answer 1

Use $u$-substitution.

Let $u = g(x)$. Then $g^{\prime}(x) = \dfrac{\text{d}u}{\text{d}x}$, so we may write $\text{d}u = g^{\prime}(x) \text{ d}x$. Hence, our integral is $$\int u \text{ d}u = \dfrac{u^2}{2} + C = \dfrac{[g(x)]^2}{2}+C\text{.}$$

Answer 2

Hint. One may recall that, using the chain rule, $$ \left[(g(x))^\alpha \right]'=\alpha \cdot g'(x)\cdot(g(x))^{\alpha-1}. $$ applying it with $\alpha=2$ gives the answer.

Answer 3

Integration by parts?

$$\int g(x)(g'(x)\ dx)=g(x)g(x)-\int g'(x)(g(x)\ dx)$$

$$2\int g(x)g'(x)\ dx=(g(x))^2+c$$