b)
$\textbf{Cách 1}$
Ta có
$ \dfrac{x^{2}}{\sqrt{x^{2}+1}}=\dfrac{1}{2}.\dfrac{x^2}{\sqrt{x^{2}
+1}}+\dfrac{1}{2}.\dfrac{x^2+1-1}{\sqrt{x^{2}
+1}}$
$=\dfrac{1}{2} \sqrt{x^{2} +1}+\dfrac{1}{2}.\dfrac{x^2}{\sqrt{x^{2}
+1}}-\dfrac{1}{2}.\dfrac{1}{\sqrt{x^{2} +1}}$
$=\dfrac{1}{2} \sqrt{x^{2} +1}+\dfrac{1}{2}.\dfrac{x^2}{\sqrt{x^{2} +1}}-\dfrac{1}{2}.\dfrac{1+\dfrac{x}{\sqrt{x^{2}
+1}}}{x+\sqrt{x^{2} +1}}$
$=\left (\dfrac{1}{2} x\sqrt{x^{2} +1} \right )'-\left
( \dfrac{1}{2}\ln\left| {x+\sqrt{x^{2} +1}} \right| \right )'$
vậy
$ \int\limits_{\sqrt{2}}^{3}\dfrac{x^{2}}{\sqrt{x^{2}+1}}dx=\left[ {\dfrac{1}{2}
x\sqrt{x^{2} +1}- \dfrac{1}{2}\ln\left| {x+\sqrt{x^{2} +1}} \right|} \right]_{\sqrt{2}}^{3}$
Đến đây bạn tự thay số nốt vào nhé.