JEE Main 2014 · 19 Apr · Shift 1 · Mathematics Q75

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MathematicsSingle correct+4 / -1Question 75 of 90

Let function F be defined as $F(x)=\displaystyle\int_1^x\frac{e^t}t\,dt$ , $x>0$ then the value of the integral $\displaystyle\int_1^x\frac{e^t}{t+a}\,dt$ , 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)]

Answer and worked solution

Official answer: D

Put $s=t+a$ :
$\int_{1+a}^{x+a}\frac{e^{s-a}}s\,ds=e^{-a}[F(x+a)-F(1+a)].$

Topics:Properties of definite integrals

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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 75