Answer
$$\int e^{\sqrt{3s+9}}ds=\frac{2}{3}e^{\sqrt{3s+9}}(\sqrt{3s+9}-1)+C$$
Work Step by Step
$$A=\int e^{\sqrt{3s+9}}ds$$
Let $x=\sqrt{3s+9}$. We would have $dx=\frac{(3s+9)'}{2\sqrt{3s+9}}ds=\frac{3}{2\sqrt{3s+9}}ds=\frac{3}{2x}ds$
That means $ds=\frac{2x}{3}dx$
$$A=\int e^x\times\frac{2x}{3}dx=\frac{2}{3}\int xe^x dx$$
Take $u=x$ and $dv=e^x dx$
We then have $du=dx$ and $v=e^x$
Apply the formula $\int udv= uv-\int vdu$, we have $$A=\frac{2}{3}\Big(xe^x-\int e^xdx\Big)$$ $$A=\frac{2}{3}\Big(xe^x-e^x\Big)+C$$ $$A=\frac{2}{3}e^x(x-1)+C$$
Replace $x$ back with $\sqrt{3s+9}$: $$A=\frac{2}{3}e^{\sqrt{3s+9}}(\sqrt{3s+9}-1)+C$$
