Question

Let function $F$ be defined as $F\left(x\right)={\int }_1^x\frac{e^t}{t}dt,x>0$, then the value of the integral ${\int }_1^x\frac{e^t}{t+a}dt,$ where a > 0, is :

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

Answer 1

The correct option is D $e^{-a}\left[F\right(x+a)-F(1+a\left)\right]$

The function $F\left(x\right)$ is hard to integrate directly. It is easier to use Leibniz Integral rule instead.

Let $G\left(x\right)={\int }_1^x\frac{e^t}{t+a}dt$. $\therefore G^{\prime }\left(x\right)=\frac{e^x}{x+a}\frac{d\left(x\right)}{dx}-\frac{e^1}{1+a}\frac{d\left(1\right)}{dx}$

$\therefore G^{\prime }\left(x\right)=\frac{e^x}{x+a}$ with $a>1$ from the limits.

Similarly, $F^{\prime }\left(x\right)=\frac{e^x}{x}$ with $x>1$.

Also, from $F^{\prime }\left(x\right)$, we can get $F^{\prime }(x+a)=\frac{e^{x+a}}{x+a}$

$\Rightarrow e^{-a}{F}^{\prime }(x+a)=\frac{e^x}{x+a}$

On integrating from $1$ to $x$, we get

$e^{-a}F(x+a)+c={\int }_1^x\frac{e^t}{t+a}dt=G\left(x\right)$

To determine the integration constant $c$, we substitute the boundary condition $x=1$ which gives

$e^{-a}F(1+a)+c={\int }_1^1\frac{e^t}{t+a}dt$

$\Rightarrow e^{-a}F(1+a)+c=0$

$\Rightarrow c=-e^{-a}F(1+a)$.

Therefore, $G\left(x\right)=e^{-a}\left(F\right(x+a)-F(1+a\left)\right)$