Question

Solve for the value of the integral, given as \[\int {{{\text{a}}^{\text{x}}}} \left[ {{\text{log x + log a log}}\left( {\dfrac{{{{\text{x}}^{\text{x}}}}}{{{{\text{e}}^{\text{x}}}}}} \right)} \right]{\text{dx}}\]=?
$
  {\text{A}}{\text{. x}}\left( {{\text{log x - 1}}} \right) \\
  {\text{B}}{\text{. }}{{\text{a}}^{\text{x}}}.{\text{x}}\left( {{\text{log x - 1}}} \right) \\
  {\text{C}}{\text{. }}{{\text{a}}^{\text{x}}}.{\text{x}}\left( {{\text{log x + 1}}} \right) \\
  {\text{D}}{\text{. }}{{\text{a}}^{\text{x}}}.\left( {{\text{log x - 1}}} \right) \\
$

Answer 1

Hint: In order to compute the integral, we divide it into parts and use relevant formulae such as integral by-parts and formula for integral of log x and ${{\text{a}}^{\text{x}}}$ and use product rule of differentiation when necessary.

Complete step-by-step answer:
Given Data,
\[\int {{{\text{a}}^{\text{x}}}} \left[ {{\text{log x + log a log}}\left( {\dfrac{{{{\text{x}}^{\text{x}}}}}{{{{\text{e}}^{\text{x}}}}}} \right)} \right]{\text{dx}}\],
To simplify, we know for a term log ${{\text{x}}^{\text{k}}}$ can be written as, k log x, Also we know ${\text{log}}\left( {\dfrac{{\text{a}}}{{\text{b}}}} \right) = {\text{loga - logb}}$ we use it in the above equation we get
$
   = \int {{{\text{a}}^{\text{x}}}} \left[ {{\text{logx + loga}}{\text{.}}{{\left( {\log {\text{x - loge}}} \right)}^{\text{x}}}} \right]{\text{dx}} \\
   = \int {{{\text{a}}^{\text{x}}}} \left[ {{\text{logx + loga}}{\text{.x}}\left( {\log {\text{x - loge}}} \right)} \right]{\text{dx}} \\
$
Now we know log e = 1, i.e. our equation becomes
$
   = \int {{{\text{a}}^{\text{x}}}} \left[ {{\text{logx + loga}}{\text{.x}}\left( {\log {\text{x - 1}}} \right)} \right]{\text{dx}} \\
   = \int {\left[ {{{\text{a}}^{\text{x}}}{\text{logx + }}{{\text{a}}^{\text{x}}}{\text{loga}}{\text{. x}}\left( {\log {\text{x - 1}}} \right)} \right]{\text{dx}}} \\
   = \int {{{\text{a}}^{\text{x}}}{\text{logx}}} {\text{ dx + }}\int {{{\text{a}}^{\text{x}}}{\text{loga}}{\text{. x}}\left( {\log {\text{x - 1}}} \right)} {\text{ dx - - - - - - (2)}} \\
$
Now let us solve the second term of the equation,
i.e. $\int {{{\text{a}}^{\text{x}}}{\text{loga}}{\text{. x}}\left( {\log {\text{x - 1}}} \right)} {\text{ dx - - - - (1)}}$
The above equation is of the form$\int {{\text{uv}}} $, we know $\int {{\text{uv}}} $= ${\text{u}}\int {\text{v}} - \int {{\text{u'}}} \int {\text{v}} $
Comparing the formula with equation (1), we consider
u = x (log x - 1) and v = ${{\text{a}}^{\text{x}}}$log a
$ \Rightarrow {\text{u' = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{x}}\left( {{\text{logx - 1}}} \right)} \right)$
It is in the form of$\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{uv}}} \right)$, which is given as $\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{uv}}} \right)$=${\text{u}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\text{v}} \right) + {\text{v}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\text{u}} \right)$, hence the equation becomes
${\text{u' = }}\left( {{\text{log x - 1}}} \right)\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\text{x}} \right) + {\text{x}}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{log x - 1}}} \right)$
$
   \Rightarrow {\text{u' = }}\left( {{\text{logx - 1}}} \right) + {\text{x}}\left( {\dfrac{1}{{\text{x}}} - 0} \right) \\
   \Rightarrow {\text{u' = logx - 1 + 1}} \\
   \Rightarrow {\text{u' = logx}} \\
$ --- As$\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{logx}}} \right) = \dfrac{1}{{\text{x}}}$.
Now our equation (1) becomes
$\int {{{\text{a}}^{\text{x}}}{\text{loga}}{\text{. x}}\left( {\log {\text{x - 1}}} \right)} {\text{ dx}}$
= $\int {{\text{uv}}} $
$ = {\text{ x}}\left( {{\text{log x - 1}}} \right)\int {{{\text{a}}^{\text{x}}}{\text{loga}}} - \int {{\text{log x}}} \int {{{\text{a}}^{\text{x}}}\log {\text{a}}} {\text{ dx}}$
Now we know$\int {{{\text{a}}^{\text{x}}}} = \dfrac{{{{\text{a}}^{\text{x}}}}}{{{\text{log a}}}}$, hence the equation becomes
$
   = {\text{ x}}\left( {{\text{log x - 1}}} \right){\text{ }}\dfrac{{{{\text{a}}^{\text{x}}}}}{{{\text{log a}}}}{\text{.log a - }}\int {{\text{log x}}} .\dfrac{{{{\text{a}}^{\text{x}}}}}{{{\text{log a}}}}{\text{.log a}} \\
  {\text{ = x}}\left( {{\text{log x - 1}}} \right){\text{ }}{{\text{a}}^{\text{x}}}{\text{ - }}\int {{\text{log x}}} .{{\text{a}}^{\text{x}}}{\text{dx}} \\
$
Now substitute the obtained value in equation (2) we get,
$\int {{{\text{a}}^{\text{x}}}{\text{logx}}} {\text{ dx + }}\int {{{\text{a}}^{\text{x}}}{\text{loga}}{\text{. x}}\left( {\log {\text{x - 1}}} \right)} {\text{ dx}}$= $\int {{{\text{a}}^{\text{x}}}{\text{logx}}} {\text{ dx + x}}\left( {{\text{log x - 1}}} \right){\text{ }}{{\text{a}}^{\text{x}}}{\text{ - }}\int {{\text{log x}}} .{{\text{a}}^{\text{x}}}{\text{dx}}$
⟹${\text{x}}\left( {{\text{log x - 1}}} \right){\text{ }}{{\text{a}}^{\text{x}}}$
⟹${{\text{a}}^{\text{x}}}.{\text{x}}\left( {{\text{log x - 1}}} \right)$
Hence Option B is the correct answer.

Note: In order to solve this type of question the key is to perform simplification on the given equation using logarithmic identities. A good knowledge in formulas of integration and differentiation of terms like log x, ${{\text{a}}^{\text{x}}}$and similar terms is appreciated. Solving the equation by splitting it makes it simpler.