Question Number 84101 by niroj last updated on 09/Mar/20
$$\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{reduction}}\:\boldsymbol{\mathrm{formula}} \\ $$$$\:\:\int\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{ax}}} \:\boldsymbol{\mathrm{dx}} \\ $$
Answered by MJS last updated on 09/Mar/20
![∫x^n e^(ax) dx= [t=a^(n+1) x^(n+1) → dx=(dt/(a^(n+1) (n+1)x^n ))] =(1/(a^(n+1) (n+1)))∫e^t^(1/(n+1)) dt= this integral is the incomplete Gamma function =... =(1/((−1)^n a^(n+1) ))Γ(n+1, −ax) +C](https://www.tinkutara.com/question/Q84107.png)
$$\int{x}^{{n}} \mathrm{e}^{{ax}} {dx}= \\ $$$$\:\:\:\:\:\left[{t}={a}^{{n}+\mathrm{1}} {x}^{{n}+\mathrm{1}} \:\rightarrow\:{dx}=\frac{{dt}}{{a}^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right){x}^{{n}} }\right] \\ $$$$=\frac{\mathrm{1}}{{a}^{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)}\int\mathrm{e}^{{t}^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} } {dt}= \\ $$$$\mathrm{this}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{the}\:\mathrm{incomplete}\:\mathrm{Gamma}\:\mathrm{function} \\ $$$$=… \\ $$$$=\frac{\mathrm{1}}{\left(−\mathrm{1}\right)^{{n}} {a}^{{n}+\mathrm{1}} }\Gamma\left({n}+\mathrm{1},\:−{ax}\right)\:+{C} \\ $$
Answered by TANMAY PANACEA last updated on 09/Mar/20
$${I}_{{n}} =\int{x}^{{n}} {e}^{{ax}} {dx} \\ $$$$={x}^{{n}} .\frac{{e}^{{ax}} }{{a}}−\int{nx}^{{n}−\mathrm{1}} .\frac{{e}^{{ax}} }{{a}}{dx} \\ $$$$=\frac{{x}^{{n}} .{e}^{{ax}} }{{a}}−\frac{{n}}{{a}}{I}_{{n}−\mathrm{1}} \\ $$
Commented by niroj last updated on 09/Mar/20
$$\:{thank}\:\:{to}\:{all} \\ $$