Let function F be defined as F(x)= _ 1 ^ x e^ t t dt, x > 0 then the value of the…
Mathematics · JEE Main · NTA Exams — Integral Calculus
Let function F be defined as \(F(x)=\int_{1}^{x} \frac{e^{t}}{t}\)dt, x > 0 then the value of the integral \(\int_{1}^{\mathrm{x}} \frac{\mathrm{e}^{\mathrm{t}}}{\mathrm{t}+\mathrm{a}} \mathrm{dt},\) where a > 0, is:
- ea[F(x) - F(1 + a)]
- e-a[F(x + a) - F(a )]
- ea[F(x + a) - F(1 + a)]
- e-a[F(x + a) - F(1 + a)]
Answer
(D) e -a [F(x + a) - F(1 + a)]
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