Question

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

Answer 1

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

& \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{aligned}$$ $$\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).