Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ và thỏa $\int\limits_{-2}^{2}{f\left( \sqrt{{{x}^{2}}+5}-x...

Đặt $t=\sqrt{{{x}^{2}}+5}-x.$
Suy ra $t+x=\sqrt{{{x}^{2}}+5}\Leftrightarrow {{\left( t+x \right)}^{2}}={{x}^{2}}+5\Leftrightarrow {{t}^{2}}+2tx=5\Leftrightarrow x=\dfrac{5}{2t}-\dfrac{t}{2}\Rightarrow dx=\left( -\dfrac{5}{2{{t}^{2}}}-\dfrac{1}{2} \right)dt.$
Đổi cận: $x=-2\Rightarrow t=5,x=2\Rightarrow t=1.$
Khi đó $1=\int\limits_{-2}^{2}{f\left( \sqrt{{{x}^{2}}+5}-x \right)dx}=\int\limits_{5}^{1}{f\left( t \right)\left( -\dfrac{5}{2{{t}^{2}}}-\dfrac{1}{2} \right)dt}=\dfrac{1}{2}\int\limits_{1}^{5}{f\left( t \right)\left( \dfrac{5}{{{t}^{2}}}+1 \right)dt}$
$\Leftrightarrow 2=\int\limits_{1}^{5}{f\left( t \right)\left( \dfrac{5}{{{t}^{2}}}+1 \right)dt}\Leftrightarrow 2=5\int\limits_{1}^{5}{\dfrac{f\left( t \right)}{{{t}^{2}}}dt}+\int\limits_{1}^{5}{f\left( x \right)dx}\Leftrightarrow 2=5\int\limits_{1}^{5}{\dfrac{f\left( x \right)}{{{x}^{2}}}dx}+\int\limits_{1}^{5}{f\left( x \right)dx}$
$\Leftrightarrow 2=5.3+\int\limits_{1}^{5}{f\left( x \right)dx}\Leftrightarrow \int\limits_{1}^{5}{f\left( x \right)dx}=-13.$