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 &#8289; = &#8747; 1 x &#160;&#160;e t t dt ,&#160; x &#8289; &#62; 0 , then the value of the integral &#8747; 1 x &#8289; e t t + a dt , where a &#62; 0 , is

Quantrex Academy · Accepted Answer

Correct answer: D. e - a F x ⁡ + a - F 1 + a

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

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

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Correct answer

Step-by-step solution

More from Definite Integration

Let the function F be defined as F x &#8289; = &#8747; 1 x &#160;&#160;e t t dt ,&#160; x &#8289; &#62; 0 , then the value of the integral &#8747; 1 x &#8289; e t t + a dt , where a &#62; 0 , is

Options

Correct answer

Step-by-step solution

More from Definite Integration

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