Question

Evaluate the following integral \[\int{{{a}^{x}}{{e}^{x}}dx}\] .

Answer 1

Hint: First of all, assume \[u\] and \[v\] using the ILATE rule (inverse, logarithmic, algebraic, trigonometric, exponent). Now, use integration by parts formula, \[\int{uvdx=u\int{vdx-\int{\dfrac{du}{dx}}}}\left( \int{vdx} \right)dx\] to expand it. Then, use \[\dfrac{d{{e}^{x}}}{dx}={{e}^{x}}\] , \[\int{{{e}^{x}}dx}={{e}^{x}}\] , and \[\dfrac{d{{a}^{x}}}{dx}={{a}^{x}}\ln a\] to simplify it further. Assume \[I=\int{{{a}^{x}}{{e}^{x}}dx}\] . Now, solve it further and get the value of \[I\] .

Complete step-by-step solution:
According to the question, we are given an expression and we have to find its value.
The given expression = \[\int{{{a}^{x}}{{e}^{x}}dx}\] …………………………………..(1)
We can observe that the above equation can not be solved directly. That is, we need to transform it into a simpler form.
We know the formula, \[\int{uvdx=u\int{vdx-\int{\dfrac{du}{dx}}}}\left( \int{vdx} \right)dx\] …………………………………(2)
For assuming \[u\] and \[v\] we have a rule, ILATE (inverse, logarithmic, algebraic, trigonometric, exponent).
Using the above rule, we have to assume \[u\] and \[v\] in equation (1).
In equation (1), we have one algebraic term \[\left( {{a}^{x}} \right)\] and one exponential term \[\left( {{e}^{x}} \right)\] . In ILATE rule algebraic comes before exponent so, we have to assume the algebraic term as \[u\] and exponential term as \[v\] .
Here, let us assume that \[u={{a}^{x}}\] and \[v={{e}^{x}}\] ………………………………..(3)
Now, using the formula shown in equation (2) and on simplifying equation (1), we get
\[\int{{{a}^{x}}{{e}^{x}}dx}={{a}^{x}}\int{{{e}^{x}}dx-\int{\dfrac{d{{a}^{x}}}{dx}}}\left( \int{{{e}^{x}}dx} \right)dx\] ………………………………………….(4)
We know the formula that \[\dfrac{d{{e}^{x}}}{dx}={{e}^{x}}\] and \[\int{{{e}^{x}}dx}={{e}^{x}}\] ……………………………………..(5)
We also know the formula that \[\dfrac{d{{a}^{x}}}{dx}={{a}^{x}}\ln a\] ………………………………………(6)
Now, from equation (4), equation (5), and equation (6), we get
\[\int{{{a}^{x}}{{e}^{x}}dx}={{a}^{x}}{{e}^{x}}-\ln a\int{{{a}^{x}}}{{e}^{x}}dx\] ……………………………….(7)
Let us assume that \[I=\int{{{a}^{x}}{{e}^{x}}dx}\] ………………………………….(8)
Using equation (8), and on replacing \[\int{{{a}^{x}}{{e}^{x}}dx}\] by \[I\] in equation (7) , we get
\[\begin{align}
  & \Rightarrow I={{a}^{x}}{{e}^{x}}-\ln a\times I \\
 & \Rightarrow I+\ln a\times I={{a}^{x}}{{e}^{x}} \\
 & \Rightarrow I\left( 1+\ln a \right)={{a}^{x}}{{e}^{x}} \\
\end{align}\]
\[\Rightarrow I=\dfrac{{{a}^{x}}{{e}^{x}}}{\left( 1+\ln a \right)}\] …………………………………(9)
Now, from equation (8) and equation (9), we get
\[\therefore \int{{{a}^{x}}{{e}^{x}}dx}=\dfrac{{{a}^{x}}{{e}^{x}}}{\left( 1+\ln a \right)}\]
Hence, the value of the expression \[\int{{{a}^{x}}{{e}^{x}}dx}\] is \[\dfrac{{{a}^{x}}{{e}^{x}}}{\left( 1+\ln a \right)}\].

Note: For this type of question where we have to integrate an expression containing inverse, logarithmic, algebraic, trigonometric, and exponential terms. Always use integral by part formula, \[\int{uvdx=u\int{vdx-\int{\dfrac{du}{dx}}}}\left( \int{vdx} \right)dx\] after assuming \[u\] and \[v\] using ILATE rule(inverse, logarithmic, algebraic, trigonometric, exponent).