## Calculus 8th Edition

$$\int x^2\sqrt{2+x}dx=\frac{2}{105}(2+x)^{3/2}(15x^2-24x+32)+c$$
To solve the integral $\int x^2\sqrt{2+x}dx$ we will use the substitution $t=2+x$ which gives us $dt=dx,x=t-2.$ Putting this into the integral we get: $$\int x^2\sqrt{2+x}dx=\int (t-2)^2\sqrt{t}dt=\int(t^2-4t+4)t^{1/2}dt=\int t^{5/2}dt-4\int t^{3/2}dt+4\int t^{1/2}dt= \frac{t^{7/2}}{\frac{7}{2}}-4\frac{t^{5/2}}{\frac{5}{2}}+4\frac{t^{3/2}}{\frac{3}{2}}+c=\frac{2}{7}t^{7/2}-\frac{8}{5}t^{5/2}+\frac{8}{3}t^{3/2}+c=2t^{3/2}(\frac{1}{7}t^2-\frac{4}{5}t+\frac{4}{3})+c$$ where $c$ is arbitrary constant. Now we have to express our solution in terms of $x$ by expressing $t$ in terms of $x$: $$\int x^2\sqrt{2+x}dx=2t^{3/2}(\frac{1}{7}t^2-\frac{4}{5}t+\frac{4}{3})+c=2(2+x)^{3/2}(\frac{1}{7}(2+x)^2-\frac{4}{5}(2+x)+\frac{4}{3})+c=2(2+x)^{3/2}\frac{15(4+4x+x^2)-84(2+x)+140}{105}+c=\frac{2}{105}(2+x)^{3/2}(15x^2-24x+32)+c$$