Cho hàm số f(x) liên tục trên [-1; 1] thỏa mãn fx−1=∫−11x+etftdt. Tích phân I=∫−11exfxdx bằng:

Ta có:

fx−1=∫−11x+etftdt⇔fx−1=x∫−11ftdt+∫−11etftdt *

Giả sử ∫−11ftdt=a,∫−11etftdt=b⇒fx−1=xa+b⇔fx=ax+b+1.

Thay vào (*) ta có:

ax+b=x∫−11at+b+1dt+∫−11etat+b+1dt

⇔ax+b=xat22+bt+t1−1+∫−11etat+b+1dt

⇔ax+b=xa2+b+1−a2+b+1+at+b+1et−11−a∫−11etdt

⇔ax+b=x2b+2+a+b+1e−−a+b+1e−1−ae−e−1

⇔ax+b=x2b+2+b+1e+2a−b−1e−1

⇒a=2b+2b=b+1e+2a−b−1e−1⇔a=2b+2b=b+1e+3b+3e−1

⇔a=2b+2b=e+3eb+e+3e⇔a=2e2e2−e−1b=e+3e1−e−3e=e2+3−e2+e−3

Vậy ∫−11etftdt=∫−11exfxdx=e2+3−e2+e−3=I.

Chọn C.