Cho $\Large I=\int_1^{5}f(x)dx=26$. Khi đó $\Large J=\int_0^{2}x.[f(x^

Cho $\Large I=\int_1^{5}f(x)dx=26$. Khi đó $\Large J=\int_0^{2}x.[f(x^{2}+1)+1]dx$ bằng

• A. 15
• B. 52
• C. 54
• D. 15

Đặt $\Large u=x^{2}+1\Rightarrow du=2xdx$

Khi đó $\Large J=\dfrac{1}{2}\int_1^{5}[f(u)+1]du=\dfrac{1}{2}\left(\int_1^{5}f(u)du+\int_1^{5}1du\right)=\dfrac{1}{2}(26+x|_1^{5})=\dfrac{1}{2}(26+5-1)=15$