Question

The value of $ \int {{5^{{{\log }_e}x}}dx} $ is
A) $ \dfrac{{{x^{{{\log }_e}5 + 1}}}}{{{{\log }_e}5 + 1}} $
B) $ {5^{{{\log }_e}x}} + c $
C) $ \dfrac{{{5^{{{\log }_e}x}} + 1}}{{{{\log }_e}x + 1}} $
D) None of these

Answer 1

Hint: Write the constant (5) as $ {e^{{{\log }_e}5}} $ as the e, Euler’s constant, and the logarithm gets cancelled. Then apply integration.

Complete step-by-step answer:
We are given to find the integral of $ \int {{5^{{{\log }_e}x}}dx} $
 $
  \int {{5^{{{\log }_e}x}}dx} \\
  5 = {e^{{{\log }_e}5}} \\
   = \int {{{\left( {{e^{{{\log }_e}5}}} \right)}^{{{\log }_e}x}}dx} \\
   = \int {{e^{{{\log }_e}5\left( {{{\log }_e}x} \right)}}} dx \\
  \left( {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right) \\
   = \int {{{\left( {{e^{{{\log }_e}x}}} \right)}^{{{\log }_e}5}}dx} \\
  \left( {\because {{\left( {{a^m}} \right)}^n} = {{\left( {{a^n}} \right)}^m}} \right) \\
  {\log _e}x = \ln x \\
   \to \int {{{\left( {{e^{\ln x}}} \right)}^{\ln 5}}dx} \\
  {e^{\ln x}} = x \\
   \to \int {{x^{\ln 5}}dx} \\
  $
Integration of x power n is x power n+1 divided by n+1 $ \to \int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
 $
   \to \dfrac{{{x^{\ln 5 + 1}}}}{{\ln 5 + 1}} \\
   \to \dfrac{{{x^{{{\log }_e}5 + 1}}}}{{{{\log }_e}5 + 1}} \\
  $
The value of $ \int {{5^{{{\log }_e}x}}dx} $ is $ \dfrac{{{x^{{{\log }_e}5 + 1}}}}{{{{\log }_e}5 + 1}} $
So, the correct answer is “Option A”.

Note: Differentiation represents the rate of change of a function. Integration represents an accumulation or sum of a function over a range. Differentiation and Integration both can have limits. They both are literally inverses. So, do not confuse differentiation with integration. The natural logarithm of x is the power to which we would have to be raised to equal x. e raised to the power of logarithm of x will always result in x.