## Calculus 8th Edition

$$\int_{0}^ax\sqrt{x^2+a^2}dx=\frac{1}{3}a^3(2\sqrt{2}-1).$$
To solve this integral we will use the substitution $x^2+a^2=t$ giving $dt=t'dx=(x^2+a^2)'dx=2xdx$. Now we have $xdx=\frac{1}{2}dt.$ For the bounds: $x_1=0\to t_1=0^2+a^2=a^2$ $x_2=a \to t_2=a^2+a^2=2a^2$ Now the integral becomes $$\int_{0}^ax\sqrt{x^2+a^2}dx=\frac{1}{2}\int_{a^2}^{2a^2}\sqrt{t} dt=\frac{1}{2}\int_{a^2}^{2a^2}t^\frac{1}{2}dt=\frac{1}{2}\left.\frac{t^\frac{3}{2}}{\frac{3}{2}}\right|_{t=a^2}^{t=2a^2}=\frac{1}{3}\bigg(\Big(2a^2\Big)^\frac{3}{2}-\Big(a^2\Big)^\frac{3}{2}\bigg)=\frac{1}{3}\big(2\sqrt{2}a^3-a^3\big)=\frac{1}{3}a^3(2\sqrt{2}-1).$$