g ( r ) = ∫ 0 r x 2 + 4 d x {\displaystyle g(r)=\int _{0}^{r}{\sqrt {x^{2}+4}}\,dx}
d d r ( g ( r ) ) = d d r [ ∫ 0 r x 2 + 4 d x ] = ( 1 ) ⋅ ( r ) 2 + 4 − ( 0 ) ⋅ ( 0 ) 2 + 4 = r 2 + 4 {\displaystyle {\frac {d}{dr}}(g(r))={\frac {d}{dr}}\left[\int _{0}^{r}{\sqrt {x^{2}+4}}\,dx\right]=(1)\cdot {\sqrt {(r)^{2}+4}}-(0)\cdot {\sqrt {(0)^{2}+4}}={\sqrt {r^{2}+4}}}
Therefore, g ′ ( r ) = r 2 + 4 {\displaystyle {\text{Therefore, }}g'(r)={\sqrt {r^{2}+4}}}
![{\displaystyle {\frac {d}{dr}}(g(r))={\frac {d}{dr}}\left[\int _{0}^{r}{\sqrt {x^{2}+4}}\,dx\right]=(1)\cdot {\sqrt {(r)^{2}+4}}-(0)\cdot {\sqrt {(0)^{2}+4}}={\sqrt {r^{2}+4}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/19a27b2aeb03a9b4e43b9eb01e532c47576d3cf5)
