## Calculus 8th Edition

$$\int x\sqrt{x+2} dx=2(x+2)^{3/2}\frac{3x-4}{15}+c$$
To evaluate the integral $$\int x\sqrt{x+2} dx$$ we will use substitution $x+2=t$ which gives us $dx=dt$ and $x=t-2,$ so we get: $$\int x\sqrt{x+2} dx=\int (t-2)\sqrt t dt=\int t^{3/2}dt-2\int t^{1/2}dt= \frac{t^{5/2}}{\frac{5}{2}}-2\frac{t^{3/2}}{\frac{3}{2}}+c= \frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2}+c=2t^{3/2}\left(\frac{t}{5}-\frac{2}{3}\right)+c,$$ where $c$ is arbitrary constant. Now we have to express solution in terms of $x$: $$\int x\sqrt{x+2} dx=2t^{3/2}\left(\frac{t}{5}-\frac{2}{3}\right)+c= 2(x+2)^{3/2}\left(\frac{x+2}{5}-\frac{2}{3}\right)+c= 2(x+2)^{3/2}\frac{3(x+2)-10}{15}+c=2(x+2)^{3/2}\frac{3x-4}{15}+c$$