Question

integrate this and show with steps ​

Answer 1

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

$\displaystyle \:\bf :\longmapsto\: \int \: {2}^{ {2}^{ {2}^{x} } } {2}^{ {2}^{x} } {2}^{x} dx$

Here we use method of Substitution

$\red{\rm :\longmapsto\:Put \: {2}^{ {2}^{ {2}^{x} } } = y}$

On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto\:\dfrac{d}{dx}\: {2}^{ {2}^{ {2}^{x} } } =\dfrac{d}{dx} \: y}$

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

$\green{\boxed{ \because \: \bf \: \dfrac{d}{dx} {a}^{x} \: = \: {a}^{x}loga}}$

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

$\red{\rm :\longmapsto\: {2}^{ {2}^{ {2}^{x} } } \: {2}^{ {2}^{x} } \: {(log2)}^{2} \dfrac{d}{dx} {2}^{ {x} } = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\: {2}^{ {2}^{ {2}^{x} } } \: {2}^{ {2}^{x} } \: {(log2)}^{2} \: {2}^{ {x} } \: log(2) = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\: {2}^{ {2}^{ {2}^{x} } } \: {2}^{ {2}^{x} } \: {2}^{x} \: {(log2)}^{3} \: = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\: {2}^{ {2}^{ {2}^{x} } } \: {2}^{ {2}^{x} } \: {2}^{x} \: {(log2)}^{3} \:dx \: = dy}$

$\red{\rm :\longmapsto\: {2}^{ {2}^{ {2}^{x} } } \: {2}^{ {2}^{x} } \: {2}^{x} \: \:dx \: = \dfrac{dy}{ {(log2)}^{3} } }$

Thus,

The given integral can be rewritten as

$\displaystyle \:\bf :\longmapsto\: \int \: {2}^{ {2}^{ {2}^{x} } } {2}^{ {2}^{x} } {2}^{x} dx$

$\displaystyle \: \rm \: = \: \: \int \: \: \dfrac{dy}{ {(log2)}^{3} }$

$\displaystyle \:\rm \: = \: \: \dfrac{1}{ {(log2)}^{3} } \: \: \int \: \: dy$

$\displaystyle \:\rm \: = \: \: \dfrac{1}{ {(log2)}^{3} } \: \: y + c$

$\displaystyle \:\rm \: = \: \: \dfrac{1}{ {(log2)}^{3} } \: \: {2}^{ {2}^{ {2}^{x} } } + c$

Additional Information :-

$\green{\boxed{ \bf \: \int \:k dx = kx+ c}}$

$\green{\boxed{ \bf \: \int \: {e}^{x} dx = {e}^{x} + c}}$

$\green{\boxed{ \bf \: \int \: {a}^{x} dx = {a}^{x} log(a) + c}}$

$\green{\boxed{ \bf \: \int \: \frac{1}{x} dx =logx + c}}$

$\green{\boxed{ \bf \: \int \: sinxdx = - cosx + c}}$

$\green{\boxed{ \bf \: \int \: cosxdx = sinx+ c}}$

$\green{\boxed{ \bf \: \int \: secx \: tanxdx = secx+ c}}$

$\green{\boxed{ \bf \: \int \:cosecx \: cotx dx = - \: cosecx \: + c}}$