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Answer 1

Question :- Evaluate the following integral

$\displaystyle\int\rm {5}^{ {5}^{ {5}^{x}}} {5}^{ {5}^{x}}{5}^{x} \: dx$

$\purple{\large\underline{\sf{Solution-}}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm {5}^{ {5}^{ {5}^{x}}} {5}^{ {5}^{x}}{5}^{x} \: dx$

To evaluate this integral, we use Method of Substitution

So, Substitute

$\red{\rm :\longmapsto\: {5}^{ {5}^{ {5}^{x}}} = y}$

So, on

$\red{\rm :\longmapsto\:{5}^{ {5}^{ {5}^{x}}}log5\dfrac{d}{dx}{5}^{ {5}^{x}} = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\:{5}^{ {5}^{ {5}^{x}}}{5}^{ {5}^{x}} {(log5)}^{2} \dfrac{d}{dx}{5}^{x} = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\:{5}^{ {5}^{ {5}^{x}}}{5}^{ {5}^{x}}{5}^{x} {(log5)}^{3} dx = dy}$

$\red{\rm :\longmapsto\:{5}^{ {5}^{ {5}^{x}}}{5}^{ {5}^{x}}{5}^{x}dx = \dfrac{dy}{ {(log5)}^{3} }}$

So, on substituting these values in given integral, we get

$\rm :\longmapsto\:\displaystyle\int\rm {5}^{ {5}^{ {5}^{x}}}{5}^{ {5}^{x}}{5}^{x} \: dx$

$\rm \: = \: \displaystyle\int\rm \frac{dy}{ {(log5)}^{3} }$

$\rm \: = \: \dfrac{1}{ {(log5)}^{3} }\displaystyle\int\rm dy$

$\rm \: = \: \dfrac{1}{ {(log5)}^{3} } \: y \: + \: c$

$\rm \: = \: \dfrac{{5}^{ {5}^{ {5}^{x}}}}{ {(log5)}^{3} } \: \: + \: c$

Hence,

$\: \: \: \ \: \\ \: \: \: \: \: \: \: \: \: \boxed{ \tt{ \: \: \displaystyle\int\rm {5}^{ {5}^{ {5}^{x}}}{5}^{ {5}^{x}}{5}^{x}dx = \: \dfrac{{5}^{ {5}^{ {5}^{x}}}}{ {(log5)}^{3} } \: \: + \: c \: }}$

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More to know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Question :- Evaluate the following integral

$\displaystyle\int\rm {5}^{ {5}^{ {5}^{x}}} {5}^{ {5}^{x}}{5}^{x} \: dx$

$\purple{\large\underline{\sf{Solution-}}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm {5}^{ {5}^{ {5}^{x}}} {5}^{ {5}^{x}}{5}^{x} \: dx$

To evaluate this integral, we use Method of Substitution

So, Substitute

$\red{\rm :\longmapsto\: {5}^{ {5}^{ {5}^{x}}} = y}$

So, on

$\red{\rm :\longmapsto\:{5}^{ {5}^{ {5}^{x}}}log5\dfrac{d}{dx}{5}^{ {5}^{x}} = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\:{5}^{ {5}^{ {5}^{x}}}{5}^{ {5}^{x}} {(log5)}^{2} \dfrac{d}{dx}{5}^{x} = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\:{5}^{ {5}^{ {5}^{x}}}{5}^{ {5}^{x}}{5}^{x} {(log5)}^{3} dx = dy}$

$\red{\rm :\longmapsto\:{5}^{ {5}^{ {5}^{x}}}{5}^{ {5}^{x}}{5}^{x}dx = \dfrac{dy}{ {(log5)}^{3} }}$

So, on substituting these values in given integral, we get

$\rm :\longmapsto\:\displaystyle\int\rm {5}^{ {5}^{ {5}^{x}}}{5}^{ {5}^{x}}{5}^{x} \: dx$

$\rm \: = \: \displaystyle\int\rm \frac{dy}{ {(log5)}^{3} }$

$\rm \: = \: \dfrac{1}{ {(log5)}^{3} }\displaystyle\int\rm dy$

$\rm \: = \: \dfrac{1}{ {(log5)}^{3} } \: y \: + \: c$

$\rm \: = \: \dfrac{{5}^{ {5}^{ {5}^{x}}}}{ {(log5)}^{3} } \: \: + \: c$

Hence,

$\: \: \: \ \: \\ \: \: \: \: \: \: \: \: \: \boxed{ \tt{ \: \: \displaystyle\int\rm {5}^{ {5}^{ {5}^{x}}}{5}^{ {5}^{x}}{5}^{x}dx = \: \dfrac{{5}^{ {5}^{ {5}^{x}}}}{ {(log5)}^{3} } \: \: + \: c \: }}$

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More to know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$