Question

Evaluate $y = \int {{e^{3a.\log x}} + {e^{3x.\log a}}dx} $

Answer 1

Hint: In order to define the integral of the above expression, separate the integral into both the terms .Now use the property of logarithm \[n\log m = \log {m^n}\] to rewrite both the terms .Now considering the fact that the exponent and log are basically the inverse of each other so \[{e^{\log (n)}} = n\] . Rewrite both the integral using this property .Now to calculate the integral of first term use the rule $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$and for the second term use the integration by substitution method by substituting $u = 3x$and use the rule $\int {{a^x}dx} = \dfrac{{{a^x}}}{{\ln a}} + C$to get your final answer.

Complete step-by-step answer:
We are given an indefinite integral $y = \int {{e^{3a.\log x}} + {e^{3x.\log a}}dx} $
 \[I = y = \int {{e^{3a.\log x}} + {e^{3x.\log a}}dx} \]
Integration separate into the term, so
 \[I = y = \int {{e^{3a.\log x}}dx} + \int {{e^{3x.\log a}}dx} \]
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
But we also need to know that the number $e$and $\log $are actually inverses of each other.
First, we are going to rewrite the terms with the help of the following properties of natural logarithms.
 \[
  n\log m = \log {m^n} \\
  {e^{\log (n)}} = n \\
 \]
So, \[{e^{3a.\log x}}dx\] and \[{e^{3x.\log a}}dx\] can be written as
 \[{e^{3a.\log x}} = {e^{\log {x^{3a}}}}\] and \[{e^{3x.\log a}} = {e^{\log {a^{3x}}}}\]
Putting all these into original integration, we get
 \[I = y = \int {{e^{\log {x^{3a}}}}dx} + \int {{e^{\log {a^{3x}}}}dx} \]
Now applying the rule that \[{e^{\log (n)}} = n\]
 \[I = \int {{x^{3a}}dx} + \int {{a^{3x}}dx} \]
And we know the rule of integration$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$in the first term and use the substitution method in second integral by substituting $u = 3x \to \dfrac{{du}}{{dx}} = 3\, \to dx = \dfrac{1}{3}du$
 \[I = \dfrac{{{x^{3a + 1}}}}{{3a + 1}} + \dfrac{1}{3}\int {{a^u}du} \]
Using the rule of integral in the second term $\int {{a^x}dx} = \dfrac{{{a^x}}}{{\ln a}} + C$where C is the constant of integration
 \[I = \dfrac{{{x^{3a + 1}}}}{{3a + 1}} + \dfrac{{{a^{3x}}}}{{3\ln (a)}} + C\]
Therefore, the integral of \[y\] is equal to \[\dfrac{{{x^{3a + 1}}}}{{3a + 1}} + \dfrac{{{a^{3x}}}}{{3\ln (a)}} + C\] where C is the
So, the correct answer is “ \[\dfrac{{{x^{3a + 1}}}}{{3a + 1}} + \dfrac{{{a^{3x}}}}{{3\ln (a)}} + C\] ”.

Note: 1.Different types of methods of Integration:
A.Integration by Substitution
B.Integration by parts
C.Integration of rational algebraic function by using partial fraction
3. Value of the constant ”e” is equal to 2.71828.
4. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number , we actually undo an exponentiation.