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 If In=xneαxdx for n1, then αIn+nIn1 isequal to 

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a
xneαx+C
b
xn+C
c
eαx+C
d
xn+eαx+C

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detailed solution

Correct option is A

In=∫xneαxdx=xneαxα−∫nxn−1eαxαdx Using ∫fgdx=f∫gdx−∫f1∫gdxdxIn=xnαeαx−nα∫xn−1eαxdxαIn=xneαx−nIn−1αIn+nIn−1=xneαx+C


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