Question

Choose the correct option $\int{\left[ {{e}^{a\log x}}+{{e}^{x\log a}} \right]}dx=\_\_\_\_+x;a, x > 1$\[\]
A. $\dfrac{{{x}^{a+1}}}{a+1}+\dfrac{{{a}^{x}}}{\log a}$\[\]
B.$\dfrac{{{e}^{a\log x}}}{a\log x}+\dfrac{{{e}^{x\log a}}}{x\log a}$\[\]
C. $\dfrac{{{x}^{a-1}}}{a-1}+{{a}^{x}}\log a$\[\]
D.$\dfrac{{{e}^{a\log x}}}{\dfrac{a}{x}}+\dfrac{{{e}^{x\log a}}}{x}$\[\]

Answer 1

Hint: We denote the integral as \[I=\int{\left[ {{e}^{a\log x}}+{{e}^{x\log a}} \right]}dx=\int{{{e}^{a\log x}}}dx+\int{{{e}^{x\log a}}dx}={{I}_{1}}+{{I}_{2}}\]. We find ${{I}_{1}}$ using the standard integral on power on $x.$ We find ${{I}_{2}}$ by substituting $u=x\log a$ and the integrating with respect to $u.$ We use the logarithmic identities $m\log n=\log {{n}^{m}}$ and ${{e}^{\log \left( f\left( x \right) \right)}}=f\left( x \right)$ where they are necessary.

Complete step-by-step answer:
We know the standard integration for power on $x$ as $\int{{{x}^{n}}}=\dfrac{{{x}^{n+1}}}{n+1}+c,n\in R$ and exponential function $\int{{{e}^{x}}}={{e}^{x}}+c$ where $c$ is any constant of integration.

Let us denote the given integral to evaluate in the question as $I.$ We have
\[I=\int{\left[ {{e}^{a\log x}}+{{e}^{x\log a}} \right]}dx\]
We are also given the condition $a,x > 1$ so that logarithms are well-defined. We know from logarithm that for some real numbers $m > 0,n > 1$ that $m\log n=\log {{n}^{m}}$. We use this identity in the integrand and proceed to have,
\[\Rightarrow I=\int{\left[ {{e}^{\log {{x}^{a}}}}+{{e}^{\log {{a}^{x}}}} \right]}dx\]
We know that logarithm and exponential are inverse function which gives us the result ${{e}^{\log \left( f\left( x \right) \right)}}=f\left( x \right)$ for any continuous function $f\left( x \right).$ We use it and proceed to have,
\[\Rightarrow I=\int{\left[ {{x}^{a}}+{{a}^{x}} \right]}dx\]
We use the rule of sum of two functions in the integrand and have.
\[\Rightarrow I=\int{{{x}^{a}}dx}+\int{{{a}^{x}}dx}={{I}_{1}}+{{I}_{2}}\left( \text{say} \right)....\left( 1 \right)\]
Let us evaluate ${{I}_{1}}$ first using the standard integration for power on $x$ as,
${{I}_{1}}=\int{{{x}^{a}}dx}=\dfrac{{{x}^{a+1}}}{a+1}+{{c}_{1}}$
Here ${{c}_{1}}$is any real constant of integration. Let us proceed to evaluate ${{I}_{2}}$ . We use formula ${{e}^{\log \left( f\left( x \right) \right)}}=f\left( x \right)$ for \[f\left( x \right)=a\]. We have,
\[\begin{align}
  & {{I}_{2}}=\int{{{a}^{x}}dx} \\
 & \Rightarrow {{I}_{2}}={{\int{\left( {{e}^{\log a}} \right)}}^{x}} \\
\end{align}\]
Let us use the identity ${{\left( {{r}^{m}} \right)}^{n}}={{r}^{mn}}$ where $r\ne 0,m,n$are real numbers. We have
\[\Rightarrow {{I}_{2}}=\int{{{e}^{x\log a}}}dx\]
Let us assign $x\log a=u$, then we differentiate both side with respect to $u$ to get $\log a\dfrac{dx}{du}=1\Rightarrow dx=\dfrac{du}{\log a}$. We substitute in the above integral to have,
\[\Rightarrow {{I}_{2}}=\int{{{e}^{u}}\dfrac{du}{\log a}}=\dfrac{1}{\log a}\int{{{e}^{u}}}du\]
We integrate the exponential function with respect to $u$ and have,
\[\Rightarrow {{I}_{2}}=\dfrac{1}{\log a}{{e}^{u}}+{{c}_{2}}\]
Here ${{c}_{2}}$ is a real constant of integration. We put back $x\log a=u$ and have,
\[\Rightarrow {{I}_{2}}=\dfrac{1}{\log a}{{e}^{x\log a}}+{{c}_{2}}\]
We use logarithmic identities at$m\log n=\log {{n}^{m}}$for $m=x,n=a$ and ${{e}^{\log \left( f\left( x \right) \right)}}=f\left( x \right)$ for $f\left( x \right)={{a}^{x}}$ in the above step and have,
\[\begin{align}
  & \Rightarrow {{I}_{2}}=\dfrac{1}{\log a}{{e}^{\log {{a}^{x}}}}+{{c}_{2}} \\
 & \Rightarrow {{I}_{2}}=\dfrac{{{a}^{x}}}{\log a}+{{c}_{2}} \\
\end{align}\]
We put the results of ${{I}_{1}},{{I}_{2}}$ in equation (1) and have,
\[\begin{align}
  & \Rightarrow I=\int{{{x}^{a}}dx}+\int{{{a}^{x}}dx}=\dfrac{{{x}^{a+1}}}{a+1}+\dfrac{{{a}^{x}}}{\log a}+{{c}_{1}}+{{c}_{2}} \\
 & \Rightarrow I=\dfrac{{{x}^{a+1}}}{a+1}+\dfrac{{{a}^{x}}}{\log a}+c \\
\end{align}\]
Here $c={{c}_{1}}+{{c}_{2}}.$

So, the correct answer is “Option A”.

Note: We note that we use substitution of variables when we have to integrate composite functions. We must be careful of the confusion of formula $\int{\dfrac{1}{x}}dx=\log \left| x \right|+c$ and also with $\int{{{a}^{x}}}dx=\dfrac{{{a}^{x}}}{\ln a}+c$ where $a > 0,a\ne 1$. We can directly evaluate ${{I}_{2}}$ using the formula $\int{{{f}^{'}}\left( x \right){{e}^{x}}dx=f\left( x \right)+c}$