Question

Let the function F be defined as $F\left( x \right) = \int_1^x {\dfrac{{{e^t}}}{t}dt} ,x > 0$ , then the value of the integral $\int_1^x {\dfrac{{{e^t}}}{{t + a}}dt} ,$ where $a > 0$, is:
A) ${e^a}\left[ {F\left( x \right) - F\left( {1 + a} \right)} \right]$
B) ${e^{ - a}}\left[ {F\left( {x + a} \right) - F\left( a \right)} \right]$
C) ${e^a}\left[ {F\left( {x + a} \right) - F\left( {1 + a} \right)} \right]$
D) ${e^{ - a}}\left[ {F\left( {x + a} \right) - F\left( {1 + a} \right)} \right]$

Answer 1

The given question is the definite integral as the range of the integration is given.
The definite integral is calculated whenever a function $f$is given over the range $\left[ {a,b} \right]$ , where $a$ and $b$ are called limits of integration, $a$ being the lower limit and $b$ being the upper limit, is given by:
$\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)$
In other words, definite integral (integral whose limits are given) is given by upper limits minus lower limits, use this formula to solve the definite integral.
Try to change the variable of integration because it often reduces an integral to one of the fundamental integrals.
The method in which we change the variable to some other variable is called the method of substitution.
When the integrand involves some trigonometric functions, we use some well-known identities to find the integrals.
Whenever the bases of the two numbers are the same, their powers are added.

Complete step-by-step answer:
Step 1: Given that:
$F\left( x \right) = \int_1^x {\dfrac{{{e^t}}}{t}dt}$
To find: $\int_1^x {\dfrac{{{e^t}}}{{t + a}}dt}$ where $a > 0$
Step 2: Make changes in the integral according to the option, as you can see option has either ${e^a}$ or ${e^{ - a}}$
We know ${e^a} \times {e^{ - a}} = 1$ , therefore, multiplying by ${e^a} \cdot {e^{ - a}}$ will not affect the number.
Thus, multiply the integral by ${e^a} \cdot {e^{ - a}}$ , we get:
$\int_1^x {\dfrac{{\;{e^a}{e^{ - a}}{e^t}}}{{t + a}}dt}$
Combine the terms ${e^a}$ and ${e^t}$, their bases are the same, so their powers are added.
$\Rightarrow \int_1^x {\dfrac{{\;{e^{ - a}}{e^{t + a}}}}{{t + a}}dt}$
Take $t + a = z$
Differentiating both sides.
$dt = dz$
Step 3: Change of limits:
As $t \to 1$
$z = t + a \\\ \because z \to 1 + a \\\$
As $t \to x$
$z = t + a \\\ \because z \to x + a \\\$
On substituting $t + a,{\text{ }}dt$, and changing the limits of the given integral, we get
$\Rightarrow \int_{1 + a}^{x + a} {\dfrac{{{e^{ - a}}{e^z}}}{z}} dz$
${e^{ - a}}$ is a constant, then we can write it outside the integral.
$\Rightarrow {e^{ - a}}\int_{1 + a}^{x + a} {\dfrac{{{e^z}}}{z}} dz$
Using the property of definite integral:
$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx$ , provided $c \in \left( {a,b} \right)$
We know that 1 lies in between the range of limits i.e. $1 \in \left( {1 + a,x + a} \right)$ , therefore,
$\Rightarrow {e^{ - a}}\int_{1 + a}^{x + a} {\dfrac{{{e^z}}}{z}} dz = {e^{ - a}}\int_{1 + a}^1 {\dfrac{{{e^z}}}{z}} dz + {e^{ - a}}\int_1^{x + a} {\dfrac{{{e^z}}}{z}} dz$
Taking ${e^{ - a}}$ as common.
$\Rightarrow {e^{ - a}}\left[ {\int_{1 + a}^1 {\dfrac{{{e^z}}}{z}} dz + \int_1^{x + a} {\dfrac{{{e^z}}}{z}} dz} \right]$
When limits of integration are interchanged, the integral is multiplied by . This a property of definite integral:
$\int\limits_a^b {f\left( x \right)} dx = - \int\limits_b^a {f\left( x \right)} dx$
Thus changing the limits of the first integral term, $\int_{1 + a}^1 {\dfrac{{{e^z}}}{z}} dz$
$\Rightarrow {e^{ - a}}\left[ { - \int_1^{1 + a} {\dfrac{{{e^z}}}{z}} dz + \int_1^{x + a} {\dfrac{{{e^z}}}{z}} dz} \right]$....................…… (1)
Step 4: Modify the answer equation (1) according to the given options.
The function $F\left( x \right) = \int_1^x {\dfrac{{{e^t}}}{t}dt}$, is the function of $x$ .
Put $x = 1 + a$ in function $F\left( x \right)$ , we have:
$\Rightarrow F\left( {1 + a} \right) = \int_1^{1 + a} {\dfrac{{{e^t}}}{t}dt}$
Using the property of definite integral:
$\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( t \right)dt}$ , i.e. we can change the dependent variable without changing the integral function.
In the function $F\left( {1 + a} \right)$ , dependent variable t can be changed to z, thus integral can also be written as:
$\Rightarrow F\left( {1 + a} \right) = \int_1^{1 + a} {\dfrac{{{e^z}}}{z}dz}$ ………………….…… (2)
Put $x = x + a$ in function $F\left( x \right)$ , we have
$\Rightarrow F\left( {x + a} \right) = \int_1^{x + a} {\dfrac{{{e^t}}}{t}dt}$
Here also using the property of definite integral:
$\int_a^b {f\left( x \right)dx} = \int_a^b {f\left( t \right)dt}$ , i.e. we can change the dependent variable without changing the integral function.
In the function $F\left( {x + a} \right)$ , dependent variable t can be changed to z, thus integral can also be written as:
$\Rightarrow F\left( {x + a} \right) = \int_1^{x + a} {\dfrac{{{e^z}}}{z}dz}$ ……………………...…… (3)
Substituting equation (2) and (3) in integral equation (1)
$\Rightarrow {e^{ - a}}\left[ { - F\left( {1 + a} \right) + F\left( {x + a} \right)} \right]$
Or ${e^{ - a}}\left[ {F\left( {x + a} \right) - F\left( {1 + a} \right)} \right]$
Final answer: On solving integral $\int_1^x {\dfrac{{{e^t}}}{{t + a}}dt}$ comes out to be ${e^{ - a}}\left[ {F\left( {x + a} \right) - F\left( {1 + a} \right)} \right]$ .

Thus the correct option is (D).

Note: In the integral calculus, we are to find a function whose differential is given. Thus, integration is a process that is the inverse of differentiation.
Let $\dfrac{{dF\left( x \right)}}{{dx}} = f\left( x \right)$, then $\smallint f\left( x \right)dx = F\left( x \right) + C$ , where C is the integration constant.
Carefully do the calculation, emphasize on each and every step especially while changing
limits of the integral.
The following are some important properties of definite integral, those will be useful in calculation definite integral more easily.
$\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx$
$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)dx}$
$\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)dx}$
$\int\limits_0^{2a} {f\left( x \right)} dx = \int\limits_0^a {f\left( x \right)} dx + \int\limits_0^a {f\left( {2a - x} \right)} dx$