Cho hàm số f(x) liên tục trên R và thỏa

Chọn B.

Đặt $t=\sqrt{x^2+5}-x.$

Suy ra $t+x=\sqrt{x^2+5}\iff {\left(t+x\right)}^2=x^2+5\iff t^2+2tx=5\iff x=\frac{5}{2t}-\frac{t}{2}\Rightarrow dx=\left(-\frac{5}{2t^2}-\frac{1}{2}\right)dt.$

Đổi cận: $x=-2\Rightarrow t=5,x=2\Rightarrow t=1.$

Khi đó $1=\underset{-2}{\overset{2}{\int }}f\left(\sqrt{x^2+5}-x\right)dx=\underset{5}{\overset{1}{\int }}f\left(t\right)\left(-\frac{5}{2t^2}-\frac{1}{2}\right)dt=\frac{1}{2}\underset{1}{\overset{5}{\int }}f\left(t\right)\left(\frac{5}{t^2}+1\right)dt$

$\iff 2=\underset{1}{\overset{5}{\int }}f\left(t\right)\left(\frac{5}{t^2}+1\right)dt\iff 2=5\underset{1}{\overset{5}{\int }}\frac{f\left(t\right)}{t^2}dt+\underset{1}{\overset{5}{\int }}f\left(x\right)dx\iff 2=5\underset{1}{\overset{5}{\int }}\frac{f\left(x\right)}{x^2}dx+\underset{1}{\overset{5}{\int }}f\left(x\right)dx$

$\iff 2=5.3+\underset{1}{\overset{5}{\int }}f\left(x\right)dx\iff \underset{1}{\overset{5}{\int }}f\left(x\right)dx=-13.$