Question

Solve $I = \int {\left( {{e^{x\log a}} + {e^{a\log x}} + {e^{a\log a}}} \right)} dx$.

Answer 1

First of all, we will separate these three terms and then integrate them individually, as we know \int {\left( {a + b} \right)} dx = \int a $$$$dx + \int b $$$$dx. After separating we will use the logarithmic properties to simplify the terms. The Logarithmic Properties we will use are:
$a\log b = \log {b^a}$ and ${e^{\log a}} = a$. We usually take the base of log to be $e$ if it is not given. Using these properties and simplifying, we will integrate the three terms individually and then add them.
Integration Formulas to be used are:
$\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$, where $c$is a constant
$\int {{a^x}} dx = \dfrac{{{a^x}}}{{\log a}} + c$, where $a$ and $c$are constants
\int a $$$$dx $= ax + c$, where $a$ and $c$are constants

Complete step by step answer:
We have $I = \int {\left( {{e^{x\log a}} + {e^{a\log x}} + {e^{a\log a}}} \right)} dx$
Using \int {\left( {a + b} \right)} dx = \int a $$$$dx$$$$ + \int b $$$$dx, we get
$I = \int {\left( {{e^{x\log a}} + {e^{a\log x}} + {e^{a\log a}}} \right)} dx = \int {{e^{x\log a}}} dx + \int {{e^{a\log x}}} dx + \int {{e^{a\log a}}} dx$
$I = \int {{e^{x\log a}}} dx + \int {{e^{a\log x}}} dx + \int {{e^{a\log a}}} dx$
Letting $\int {{e^{x\log a}}} dx = {I_1}$, $\int {{e^{a\log x}}} dx = {I_2}$ and $\int {{e^{a\log a}}} dx = {I_3}$, we get
$I = {I_1} + {I_2} + {I_3} - - - - - (1)$
First Solving ${I_1}$
${I_1} = \int {{e^{x\log a}}} dx$
Using $a\log b = \log {b^a}$ in the above expression, we get
${I_1} = \int {{e^{\log {a^x}}}} dx$
Now, using ${e^{\log a}} = a$, we get
${I_1} = \int {{a^x}} dx$
Now Using the integration formula for $\int {{a^x}} dx$,
${I_1} = \int {{a^x}} dx = \dfrac{{{a^x}}}{{\log a}} + {c_1} - - - - - (2)$, where ${c_1}$ is a constant
Solving ${I_2}$ now
${I_2} = \int {{e^{a\log x}}} dx$
Using $a\log b = \log {b^a}$ in the above equation, we get
${I_2} = \int {{e^{\log {x^a}}}} dx$
Now using ${e^{\log a}} = a$, we get
${I_2} = \int {{x^a}} dx$
We know, $\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$. So,
${I_2} = \int {{x^a}} dx = \dfrac{{{x^{a + 1}}}}{{a + 1}} + {c_2} - - - - - (3)$, where ${c_2}$ is a constant
Now, solving ${I_3}$
${I_3} = \int {{e^{a\log a}}} dx$
Using $a\log b = \log {b^a}$ we get
${I_3} = \int {{e^{\log {a^a}}}} dx$
Now using ${e^{\log a}} = a$, ${I_3}$ becomes
${I_3} = \int {{a^a}} dx$
We see that ${a^a}$ is a constant term. So,
${I_3} = \int {{a^a}} dx = x{a^a} + {c_3} - - - - - (4)$, where ${c_3}$ is a constant.
Now, using (1), (2), (3) and (4), we get
$I = \dfrac{{{a^x}}}{{\log a}} + {c_1} + \dfrac{{{x^{a + 1}}}}{{a + 1}} + {c_2} + x{a^a} + {c_3}$, where ${c_1},{c_2},{c_3}$ are constants.
Now combining three constants to one constant, we get
$I = \dfrac{{{a^x}}}{{\log a}} + \dfrac{{{x^{a + 1}}}}{{a + 1}} + x{a^a} + ({c_1} + {c_2} + {c_3})$
$I = \dfrac{{{a^x}}}{{\log a}} + \dfrac{{{x^{a + 1}}}}{{a + 1}} + x{a^a} + c$, where $c = {c_1} + {c_2} + {c_3}$.
Hence,
$I = \int {\left( {{e^{x\log a}} + {e^{a\log x}} + {e^{a\log a}}} \right)} dx = \dfrac{{{a^x}}}{{\log a}} + \dfrac{{{x^{a + 1}}}}{{a + 1}} + x{a^a} + c$, where $c$ is a constant term.

Note:
We need to be very thorough with the logarithmic properties. Also, while applying properties we should check whether we are applying the right property to the respective term or not. While Integrating, we usually forget to add the constant term but it is necessary and should be taken care of.