Answer
$$\int \sqrt{2t+1}dt=\frac{(2t+1)^{3/2}}{3}+c$$
Work Step by Step
To solve the integral $\int \sqrt{2t+1}dt$ we will use substitution $u=2t+1$ which gives us $du=2dt\Rightarrow\frac{1}{2}du=dt.$ Putting this into the integral we get:
$$\int \sqrt{2t+1}dt=\int\sqrt{u}\cdot\frac{1}{2}du=\frac{1}{2}\int u^{1/2}du=\frac{1}{2}\frac{u^{3/2}}{\frac{3}{2}}+c=\frac{1}{2}\frac{2}{3}u^{3/2}+c=\frac{u^{3/2}}{3}+c$$
where $c$ is arbitrary constant.
Now we have to express solution in terms of $t$ by expressing $u$ in terms of $t$:
$$\int \sqrt{2t+1}dt=\frac{u^{3/2}}{3}+c=\frac{(2t+1)^{3/2}}{3}+c$$
