Ta có:
$\int\limits_0^4\dfrac{x}{x+\sqrt{x^2+9}}dx$
$=\int\limits_0^4{x(\sqrt{x^2+9}-x)}{9}dx$
$=\dfrac{1}{9}\int\limits_0^4x\sqrt{x^2+9}-\dfrac{1}{9}\int\limits_0^4x^2dx$
$=\dfrac{1}{18}\int\limits_0^4\sqrt{x^2+9}d(x^2+9)-\dfrac{1}{9}\int\limits_0^4x^2dx$
$=\dfrac{(x^2+9)\sqrt{(x^2+9)}}{27}\left|\begin{array}{l}4\\0\end{array}\right.-\dfrac{x^3}{27}\left|\begin{array}{l}4\\0\end{array}\right.=\dfrac{34}{27}$