Cho hàm số y=fx thỏa mãn điều kiện ∫02f'xdxx+2=3 và f2−2f0=4. Tính tích phân I=∫01f2xdxx+12

Đáp án D

Đặt $\left\{\begin{array}{l}u=\frac{1}{x+2}\\ dv=f\text{'}\left(x\right)dx\end{array}\right.\Rightarrow \left\{\begin{array}{l}du=-\frac{1}{{\left(x+2\right)}^2}\\ v=f\left(x\right)\end{array}\right.$.

Khi đó $\underset{0}{\overset{2}{\int }}\frac{f\text{'}\left(x\right)dx}{x+2}=\frac{f\left(x\right)}{x+2}\left|{}_0^2\right.+\underset{0}{\overset{2}{\int }}\frac{f\left(x\right)dx}{{\left(x+2\right)}^2}=\frac{f\left(2\right)}{4}-\frac{f\left(0\right)}{2}+\underset{0}{\overset{2}{\int }}\frac{f\left(x\right)dx}{{\left(x+2\right)}^2}=1+\underset{0}{\overset{2}{\int }}\frac{f\left(x\right)dx}{{\left(x+2\right)}^2}$.

Suy ra $K=\underset{0}{\overset{2}{\int }}\frac{f\left(x\right)dx}{{\left(x+2\right)}^2}=2\stackrel{x=2t}{\to }K=\underset{0}{\overset{1}{\int }}\frac{f\left(2t\right)d2t}{{\left(2t+2\right)}^2}=\underset{0}{\overset{1}{\int }}\frac{f\left(2t\right)dt}{2{\left(t+1\right)}^2}=2$.

Vậy $\underset{0}{\overset{1}{\int }}\frac{f\left(2t\right)dt}{{\left(t+1\right)}^2}=4$.