Question

integrate

${e}^{x} ( {a + b {e}^{x}) }^{6} \:.dx$
​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\sf {e}^{x} \: {(a + b{e}^{x})}^{6} \: dx - - - (1)$

We use here method of Substitution,

$\red{\rm :\longmapsto\:Put \:a + b {e}^{x} = y}$

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

$\red{\rm :\longmapsto\dfrac{d}{dx} \: (a + b{e}^{x})=\dfrac{d}{dx} y}$

$\red{\rm :\longmapsto \:b {e}^{x} =\dfrac{dy}{dx}}$

$\red{\rm :\longmapsto \: {e}^{x}dx = \dfrac{dy}{b} }$

So, equation (1) can be rewritten as

$\rm :\longmapsto\:\displaystyle\int\sf \: { \pink{(a + b{e}^{x})}}^{6} \: \red{{e}^{x} \: dx}$

$\rm \: \: = \:\dfrac{1}{b} \displaystyle\int\sf {y}^{6} \: dy$

$\rm \: \: = \:\dfrac{1}{b} \: \times \dfrac{ {y}^{6 + 1} }{6 + 1} + c$

$\rm \: \: = \: \dfrac{ {y}^{7} }{7b} + c$

$\rm \: \: = \: \dfrac{ {(a + b {e}^{x}) }^{7} }{7b} + c$

Hence,

$\bf :\longmapsto\:\displaystyle\int\bf {e}^{x} \: {(a + b{e}^{x})}^{6} \: dx = \dfrac{ {(a + b{e}^{x})}^{7} }{7b} + c$

Formula Used :-

$\red{ \boxed{ \sf{ \dfrac{d}{dx}k = 0\: }}}$

$\red{ \boxed{ \sf{ \dfrac{d}{dx}{e}^{x} = {e}^{x}\: }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf {x}^{n}dx = \frac{ {x}^{n + 1} }{n + 1}+ c}}}$

Additional Information :-

$\red{ \boxed{ \sf{ \displaystyle\int\sf \dfrac{1}{x}\: dx = log(x) + c}}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf cosx \: dx = sinx + c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf sinx \: dx = - \: cosx + c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf secx \: tanx\: dx = \: secx + c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf cosecx \: cotx\: dx = - \: cosecx + c }}}$

$\red{ \boxed{ \sf{ \: \displaystyle\int\sf tanx \: dx \: = log(secx) + c }}}$

$\red{ \boxed{ \sf{ \: \displaystyle\int\sf cotx \: dx \: = log(sinx) + c }}}$

$\red{ \boxed{ \sf{ \: \displaystyle\int\sf cosecx \: dx \: = log(cosecx - cotx) + c }}}$

$\red{ \boxed{ \sf{ \: \displaystyle\int\sf secx \: dx \: = log(secx + tanx) + c }}}$