Question

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 :

• A. \(e^a\left[F(x)-F(1+a)\right]\)
• B. \(e^{-a}\left[F(x+a)-F(a)\right]\)
• C. \(e^a\left[F(x+a)-F(1+a)\right]\)
• D. \(e^{-a}\left[F(x+a)-F(1+a)\right]\)

Hint:

Substitute \(u=t+a\) so that \(e^t=e^{-a}e^u\) and the integrand becomes \(e^u/u\).

Answer 1

The Correct Option is D

Put \(u=t+a\): \(t=1\Rightarrow u=1+a\), \(t=x\Rightarrow u=x+a\), \(e^t=e^{u-a}\).
\(\int_1^x\frac{e^t}{t+a}dt=e^{-a}\int_{1+a}^{x+a}\frac{e^u}{u}du=e^{-a}\left[F(x+a)-F(1+a)\right]\), since \(F(y)-F(z)=\int_z^y\frac{e^u}{u}du\).
Answer: (4) \(e^{-a}\left[F(x+a)-F(1+a)\right]\)