Question

Solve
 $
  \int {{e^{x\log a}}{e^x}dx} = \\
  1){\text{ }}\left( {\dfrac{{{a^x}}}{{\log ae}}} \right) + c \\
  2)\;\left[ {\dfrac{{{e^x}}}{{(1 + {{\log }_e}a)}}} \right] + c \\
  3)\,{{\text{(ae)}}^x} + c \\
  4)\;\left[ {\dfrac{{{{(ae)}^x}}}{{{{\log }_e}ae}}} \right] + c \\
  5)\;\left[ {\dfrac{{{a^x}{e^x}}}{{{{\log }_x}e}}} \right] + c \\
  $

Answer 1

Hint: Here we will use the different logarithmic, and power and exponent laws to simplify the given expression and then will use the formula for the product of two terms and will simplify gradually for the required resultant value.

Complete step-by-step answer:
Take the given expression: $ I = \int {{e^{x\log a}}{e^x}dx} $
By using the identity of the logarithm, $ a\log x = \log {x^a} $ we write the above expression –
 \[I = \int {{e^{\log {a^x}}}{e^x}dx} \]
Now, again by the property “e” exponential base and log cancel each other.
 \[I = \int {{a^x}{e^x}dx} \] …. (A)
Apply integration by parts, $ \int {uvdx = u\int {vdx - \int {(u'\int {vdx)dx} } } } $
 \[I = \left[ {{e^x}.\dfrac{{.{a^x}}}{{{{\log }_e}a}} - \int {{e^x}.\dfrac{{.{a^x}}}{{{{\log }_e}a}}dx} } \right] \]
Simplify the above equation –
 $ I = \dfrac{{{e^x}.{a^x}}}{{{{\log }_e}a}} - \dfrac{1}{{{{\log }_e}a}}\int {{e^x}.{a^x}.dx} $
From the equation (A)
 $ I = \dfrac{{{e^x}.{a^x}}}{{{{\log }_e}a}} - \dfrac{1}{{{{\log }_e}a}}I $
Take “I” on the left hand side of the equation, when you move any term from one side to the opposite side then the sign of the term changes. Positive term becomes negative and vice-versa.
 $ I + \dfrac{1}{{{{\log }_e}a}}I = \dfrac{{{e^x}.{a^x}}}{{{{\log }_e}a}} $
Take Common factors common and simplify the above expression –
 $ I\left( {1 + \dfrac{1}{{{{\log }_e}a}}} \right) = \dfrac{{{e^x}.{a^x}}}{{{{\log }_e}a}} $
Take LCM (least common multiple) on the left hand side of the equation –
 $ I\left( {\dfrac{{{{\log }_e}a + 1}}{{{{\log }_e}a}}} \right) = \dfrac{{{e^x}.{a^x}}}{{{{\log }_e}a}} $
Like terms in the denominator on both the sides of the equations cancels each other.
 $ I({\log _e}a + 1) = {e^x}.{a^x} $
Term multiplicative on one side is moved to the opposite side then it goes in the numerator.
 $ I = \dfrac{{{e^x}.{a^x}}}{{({{\log }_e}a + 1)}} $
when powers are the same and bases are in multiplication then we can write as the product of the terms and its whole power.
 \[I = \dfrac{{{{(ea)}^x}}}{{({{\log }_e}a + 1)}}\]
Here we can place $ 1 = {\log _e}e $ in the above expression –
 \[I = \dfrac{{{{(ea)}^x}}}{{({{\log }_e}a + {{\log }_e}e)}}\]
When bases are same and logarithms in addition it can be written as the product of the terms –
 \[I = \dfrac{{{{(ea)}^x}}}{{({{\log }_e}ae)}}\]
Hence from the given multiple choices – fourth option is the right answer.
So, the correct answer is “Option 4”.

Note: Always remember different identities and formulas to simplify the given expression. Don’t get confused between the differentiation and the integration by parts formula. Be careful about the sign convention while moving any term from one side to the opposite side the sign of the term changes.