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Question: Evaluate the integral \(\int{{{7}^{{{7}^{{{7}^{x}}}}}}}{{.7}^{{{7}^{x}}}}{{.7}^{x}}.dx\). A. \(\df...

Evaluate the integral ∫777x.77x.7x.dx\int{{{7}^{{{7}^{{{7}^{x}}}}}}}{{.7}^{{{7}^{x}}}}{{.7}^{x}}.dx.
A. 777x(log⁡7)3+c\dfrac{{{7}^{{{7}^{{{7}^{x}}}}}}}{{{\left( \log 7 \right)}^{3}}}+c
B. 777x(log⁡7)2+c\dfrac{{{7}^{{{7}^{{{7}^{x}}}}}}}{{{\left( \log 7 \right)}^{2}}}+c
C. 777x.(log⁡7)3+c{{7}^{{{7}^{{{7}^{x}}}}}}.{{\left( \log 7 \right)}^{3}}+c
D. 777x{{7}^{{{7}^{{{7}^{x}}}}}}

Explanation

Solution

For solving this question you should know about integration of powered values. In this problem, we will solve this by assuming a term as uu and then we will find dudu for that and then put them according to the question and then we will find a new integration. We will then solve this and find the answer for this.

Complete step by step solution:
According to the question we have to find the value of ∫777x.77x.7x.dx\int{{{7}^{{{7}^{{{7}^{x}}}}}}}{{.7}^{{{7}^{x}}}}{{.7}^{x}}.dx. As we know that if a problem is given to us in a power form, then we can’t find the integration of that question directly. We have to reduce that in the simplest form which can be easily integrated. And for reducing the integrating values we will substitute any term from that as uuand this term is selected as that if we differentiate this term, then it will provide an answer which can be found for the given question. Thus we will be able to reduce this. Then we will get a new integration value which is now available for integration. So, we will now integrate this new value and by this we will get our answer.
So, according to our question: if we substitute 777x{{7}^{{{7}^{{{7}^{x}}}}}} as uu, then:
777x=u{{7}^{{{7}^{{{7}^{x}}}}}}=u
On differentiating both sides of equation, we get
d(777x)=dud\left( {{7}^{{{7}^{{{7}^{x}}}}}} \right)=du
Now by the property of chain rule, that is d(f(g(x)))dx=f′(g(x))⋅g′(x)\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=f'\left( g\left( x \right) \right)\cdot g'\left( x \right) and formula d(a)x=(a)x⋅log⁡ad{{\left( a \right)}^{x}}={{\left( a \right)}^{x}}\cdot \log a, we get

& \left( {{7}^{{{7}^{{{7}^{x}}}}}} \right)\cdot \left( {{7}^{{{7}^{x}}}} \right)\cdot \left( {{7}^{x}} \right)\cdot {{\left( \log 7 \right)}^{3}}dx=du \\\ & \Rightarrow \left( {{7}^{{{7}^{{{7}^{x}}}}}} \right)\cdot \left( {{7}^{{{7}^{x}}}} \right)\cdot \left( {{7}^{x}} \right)dx=\dfrac{du}{{{\left( \log 7 \right)}^{3}}} \\\ \end{aligned}$$ And therefore we can the given integral term as $\int{{{7}^{{{7}^{{{7}^{x}}}}}}}{{.7}^{{{7}^{x}}}}{{.7}^{x}}.dx=\int{\dfrac{du}{{{\left( \log 7 \right)}^{3}}}}$ And we can further write it as $\int{{{7}^{{{7}^{{{7}^{x}}}}}}}{{.7}^{{{7}^{x}}}}{{.7}^{x}}.dx=\dfrac{u}{{{\left( \log 7 \right)}^{3}}}$ On putting the value of u again, we get $\int{{{7}^{{{7}^{{{7}^{x}}}}}}}{{.7}^{{{7}^{x}}}}{{.7}^{x}}.dx=\dfrac{{{7}^{{{7}^{{{7}^{x}}}}}}}{{{\left( \log 7 \right)}^{3}}}$ **Hence the correct option is A.** **Note:** While solving this problem you have to be careful about selecting the value of $u$ which will be substituted. It must be in a form that when we differentiate it once then its answer must be in the given integral in any one form, otherwise it will be wrong.