Let the function F be defined as F x ⁡ = ∫ 1 x e t t dt , x ⁡ 0 , then the value of the integral ∫ 1 x ⁡ e t t + a dt , where a 0 , is

Question

Let the function F be defined as F x ⁡ = ∫ 1 x e t t dt , x ⁡ > 0 , then the value of the integral ∫ 1 x ⁡ e t t + a dt , where a > 0 , is, e a F x ⁡ - F 1 + a, e - a F x ⁡ + a - F a, e a F x ⁡ + a - F 1 + a, e - a F x ⁡ + a - F 1 + a

MARKS App · Accepted Answer

F x = ∫ 1 x e t t dt I = ∫ 1 x e t t + a dt Let, t + a = y ⇒ dt = dy Also, t = 1 ⇒ y = 1 + a and t = x ⇒ y = x + a ∴ I = ∫ 1 + a x + a e y - a y dy = e - a ∫ 1 + a x + a e y y dy = e - a ∫ 1 + a x + a e t t dt = e - a F x ⁡ + a - F 1 + a

Search any question & find its solution

Looking for more such questions to practice?