Cho hàm số f(x) liên tục trên [0;4] thỏa mãn và f(x)>0 với mọi . Biết rằng , giá trị của bằng

* Hướng dẫn giải

Đáp án A

Ta có: $f^{\text{'}\text{'}}\left(x\right).f\left(x\right)+\frac{f^2\left(x\right)}{\sqrt{{\left(2x+1\right)}^3}}={\left(f^{\text{'}}\left(x\right)\right)}^2\iff f^{\text{'}\text{'}}\left(x\right).f\left(x\right)-{\left(f^{\text{'}}\left(x\right)\right)}^2=-\frac{f^2\left(x\right)}{\sqrt{{\left(2x+1\right)}^3}}$

$\iff \frac{f^{\text{'}\text{'}}\left(x\right).f\left(x\right)-{\left(f^{\text{'}}\left(x\right)\right)}^2}{f^2\left(x\right)}=-\frac{1}{\sqrt{{\left(2x+1\right)}^3}}\iff {\left(\frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}\right)}^{\text{'}}=-\frac{1}{\sqrt{{\left(2x+1\right)}^3}}$

$\iff \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}=-\int \frac{dx}{\sqrt{{\left(2x+1\right)}^3}}\iff \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}=-\int {\left(2x+1\right)}^{-\frac{3}{2}}dx\iff \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}=\frac{1}{\sqrt{2x+1}}+C_1$

Thay $x=0$  ta được:$C_1=0\Rightarrow \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}=\frac{1}{\sqrt{2x+1}}\Rightarrow \int \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}dx=\int \frac{dx}{\sqrt{2x+1}}$

$\iff \ln \left(f\left(x\right)\right)=\sqrt{2x+1}+C_2$

Thay $x=0$  ta được:$C_2=-1\Rightarrow \ln \left(f\left(x\right)\right)=\sqrt{2x+1}-1$ .

Thay$x=4$  ta được $\ln \left(f\left(4\right)\right)=2\Rightarrow f\left(4\right)=e^2$  .