Question

Integrate the following function

$\frac{ {e}^{5logx} + {e}^{4logx}}{{e}^{3logx} + {e}^{2logx}} \\ \: with \: respect \: to \: x$
​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {e}^{5logx} + {e}^{4logx} }{{e}^{3logx} + {e}^{2logx}} \: dx$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {e}^{ylogx} \: = \: {x}^{y} \: }}}$

So, above integral can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{ {x}^{5} + {x}^{4} }{ {x}^{3} + {x}^{2} } \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{ {x}^{4} (x + 1)}{ {x}^{2} (x + 1)} \: dx$

$\rm \:  =  \: \displaystyle\int\rm {x}^{2} \: dx$

$\rm \:  =  \: \dfrac{ {x}^{2 + 1} }{2 + 1} + c$

$\rm \:  =  \: \dfrac{ {x}^{3} }{3} + c$

Hence,

$\rm\implies \:\:\boxed{\tt{ \displaystyle\int\rm \frac{ {e}^{5logx} + {e}^{4logx} }{{e}^{3logx} + {e}^{2logx}} \: dx = \frac{ {x}^{3} }{3} + c}}$

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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

Answer:

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {e}^{5logx} + {e}^{4logx} }{{e}^{3logx} + {e}^{2logx}} \: dx$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {e}^{ylogx} \: = \: {x}^{y} \: }}}$

So, above integral can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm <strong>\</strong>frac{ {x}^{5} + {x}^{4} }{ {x}^{3} + {x}^{2} } \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{ {x}^{4} (x + 1)}{ {x}^{2} (x + 1)} \: dx$

$\rm \:  =  \: \displaystyle\int\rm {x}^{2} \: dx$

$\rm \:  =  \: \dfrac{ {x}^{2 + 1} }{2 + 1} + c$

$\rm \:  =  \: \dfrac{ {x}^{3} }{3} + c$

Hence,

$\rm\implies \:\:\boxed{\tt{ \displaystyle\int\rm \frac{ {e}^{5logx} + {e}^{4logx} }{{e}^{3logx} + {e}^{2logx}} \: dx = \frac{ {x}^{3} }{3} + c}}$

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