Answer
$$\int\sqrt{2t+1}dt=\frac{1}{3}\sqrt{(2t+1)^3} + C$$
Work Step by Step
$$A=\int\sqrt{2t+1}dt$$
Let $u=2t+1$
Then $du=(2t+1)'dt=2dt$. So $dt=\frac{1}{2}du$
Substitute into $A$, we have $$A=\int \sqrt u(\frac{1}{2})du$$ $$A=\frac{1}{2}\int u^{1/2}du$$ $$A=\frac{1}{2}\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{\sqrt{u^3}}{3}+C$$ $$A=\frac{1}{3}\sqrt{(2t+1)^3} + C$$
