Question

Find The Integral
Answer with steps
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Answer 1

Step-by-step explanation:

We have,

$\rm \: I = \int {a}^{x} . {e}^{x}dx \: \: \: \: \: \: \: ....(i)$

$\rm \implies \: I = {a}^{x} \int {e}^{x}dx - \int \bigg \{ \frac{d}{dx}( {a}^{x} ). \int {e}^{x} dx \bigg \}dx$

$\rm \implies \: I = {a}^{x} . {e}^{x} - \int \{ {a}^{x} . ln(a) . {e}^{x} \}dx$

$\rm \implies \: I = {a}^{x} . {e}^{x} - ln(a) \int {a}^{x}. {e}^{x}dx$

$\rm \implies \: I = {a}^{x} . {e}^{x} - ln(a). I \: \: \: \: \: \: .... [from \: \: (i)]$

$\rm \implies \: I - ln(a). I= {a}^{x} . {e}^{x} \: \: \: \: \: \:$

$\rm \implies \: I \{1 - ln(a) \}= {a}^{x} . {e}^{x} \: \: \: \: \: \:$

$\rm \implies \: I = \frac{{a}^{x} . {e}^{x}}{1 - ln(a) } \: \: \: \: \: \:$

Answer 2

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int \rm \: {a}^{x}. {e}^{x} \: dx$

We know,

$\red{ \boxed{ \sf{ \: {x}^{n} \times {y}^{n} = {(xy)}^{n}}}}$

So, using this, the above integral can be rewritten as

$\rm \:  =  \: \displaystyle\int \rm \: {(ae)}^{x} \: dx$

We know,

$\red{ \boxed{ \sf{ \:\displaystyle\int \rm \: {a}^{x} \: dx \: = \: \frac{ {a}^{x} }{loga} + c}}}$

So, using this, we get

$\rm \:  =  \: \dfrac{ {(ae)}^{x} }{logae} \: + \: c$

Hence,

$\red{ \boxed{ \sf{ \:\displaystyle\int \rm \: {a}^{x}. {e}^{x} \: dx =  \: \dfrac{ {(ae)}^{x} }{logae} \: + \: c \: \: }}}$

Additional Information :-

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