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Question: What is \[\int {{a^x}{e^x}\,dx} \]equal to? A. \( \dfrac{{{a^x}{e^x}}}{{\ln \,a}} + \,c\, \) where...

What is ∫axex dx\int {{a^x}{e^x}\,dx} equal to?
A. axexln⁡ a+ c \dfrac{{{a^x}{e^x}}}{{\ln \,a}} + \,c\, where c is the constant of integration
B. axex{a^x}{e^x} + c where c is the constant of integration
C. axexln⁡(ae)\dfrac{{{a^x}{e^x}}}{{\ln (ae)}} + c where c is the constant of integration
D. None of the above

Explanation

Solution

Hint : n this question, we can clearly see that ∫axex dx\int {{a^x}{e^x}\,dx} is in the form of ∫u.v dx\int {u.v\,dx} . So, we can use the formula of integration of two functions or integration by parts. Choose first and second function according to the ILATE rule. Here ax{a^x} will be the first function and ex{e^x} will be the second function. Proceed with simplification after applying the product formula.

Complete step-by-step answer :
Let I = ∫axex dx\int {{a^x}{e^x}\,dx}
We know that, ∫u.vdx=u∫vdx − ∫[ddxu.∫vdx] dx\int {u.vdx = u\int {vdx\, - \,\int {\left[ {\dfrac{d}{{dx}}u.\int {vdx} } \right]} } } \,dx
Where u = ax{a^x} and v = ex{e^x} .
II = ∫ax.ex dx = ax.∫exdx − ∫[ddxax. ∫exdx]dx\int {{a^x}} .{e^x}\,dx\, = \,{a^x}.\int {{e^x}} dx\, - \,\int {\left[ {\dfrac{d}{{dx}}{a^x}.\,\int {{e^x}} dx} \right]} dx (Putting the values of u and v in the above formula)
⇒I\Rightarrow I = ax.ex− ∫ax.ln⁡a . ex dx{a^x}.{e^x} - \,\int {{a^x}} .\ln a\,.\,{e^x}\,dx (We know that ∫ex=ex\int {{e^x}} = {e^x} and ddxax=ax.ln⁡a\dfrac{d}{{dx}}{a^x} = {a^x}.\ln a )
⇒I\Rightarrow I = ax.ex − ln⁡ a∫ax.exdx{a^x}.{e^x}\, - \,\ln \,a\int {{a^x}} .{e^x}dx (ln a is constant)
⇒I\Rightarrow I = ax.ex− Iln⁡a{a^x}.{e^x} - \,I\ln a + c (Putting the value of ∫axex dx\int {{a^x}{e^x}\,dx} = I)
I+I ln⁡ a = ax.exI + I\,\ln \,a\, = \,{a^x}.{e^x} + c
I(1+ln⁡a)= axexI\left( {1 + \ln a} \right) = \,{a^x}{e^x} + c
⇒I=axex1+ln⁡ a\Rightarrow I = \dfrac{{{a^x}{e^x}}}{{1 + \ln \,a}} + c
⇒I=axexln⁡ e + ln⁡a+c\Rightarrow I = \dfrac{{{a^x}{e^x}}}{{\ln \,e\, + \,\ln a}} + c (We know that the value of ln e = 1)
⇒I=axexln⁡ae+c\Rightarrow I = \dfrac{{{a^x}{e^x}}}{{\ln ae}} + c

So, the correct answer is “Option C”.

Note : Integration by parts formula ∫u.vdx=u∫vdx − ∫[ddxu.∫vdx] dx\int {u.vdx = u\int {vdx\, - \,\int {\left[ {\dfrac{d}{{dx}}u.\int {vdx} } \right]} } } \,dx . This formula is used for integrating the product of two functions. But you should note that this formula is not applicable for functions such as ∫xcos⁡xdx\int {\sqrt x } \cos xdx . But this formula will be applicable on ∫xcos⁡x\int {x\cos x} . One more thing you need to keep in mind that we do not add any constant while finding the integral of the second function.
There is a rule of choosing the first function and the second function, that rule is called ILATE. I stands for inverse trigonometric function, L stands for logarithmic function, A stands for algebraic function, T stands for trigonometric function and E stands for exponential function. Always follow this rule while integrating by product. If you do the integration randomly, the solution becomes much more complicated.