Kkutie7
  • Kkutie7
Doing another Integration:
Mathematics
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chestercat
  • chestercat
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Kkutie7
  • Kkutie7
\[\int 3^{x}e^{x}\] I'm not sure where to start
Kkutie7
  • Kkutie7
this is what I was thinking first: \[\int e^{xln{3}}e^{x}dx\] then what u sub for xln3?
shubhamsrg
  • shubhamsrg
Have you learnt integration by parts ?

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Kkutie7
  • Kkutie7
yes I have
ganeshie8
  • ganeshie8
this should do : \(a^m\cdot a^n = a^{m+n}\)
shubhamsrg
  • shubhamsrg
That'd also do. Yes.
shubhamsrg
  • shubhamsrg
Otherwise it's a simple application of integrating by parts.
Kkutie7
  • Kkutie7
\[\int e^{xln3+x}dx\]
Kainui
  • Kainui
@shubhamsrg Integration by parts here is a bad idea
Kkutie7
  • Kkutie7
\[\frac{1}{ln(3)+1}\int e^{u}du\]
shubhamsrg
  • shubhamsrg
I = int(e^x 3^x) = e^x 3^x - int(e^x 3^x ln 3) I= e^x 3^x - I ln3 It's not too difficult to calculate I from here I guess.
Kkutie7
  • Kkutie7
answer=\[\frac{3^{x}e^{x}}{ln{3}+1}\] right?
anonymous
  • anonymous
\[\int\limits 3^x e^x dx=3^x.e^x- \int\limits3^x \ln(3) e^xdx+C\]\[(1+\ln(3))\int\limits 3^x e^x dx=3^x.e^x+C\]\[\int\limits 3^x e^x dx=\frac{3^xe^x}{1+\ln(3)}+C'\] Looks fine to me using by parts
ganeshie8
  • ganeshie8
that looks good ! below is, i think, a slightly better algebra.. \[\int 3^xe^x\,dx = \int (3e)^x\, dx = \int e^{x\ln (3e)}\, dx = \dfrac{e^{x\ln(3e)}}{\ln(3e)}+C = \dfrac{(3e)^x}{\ln(3e)}+C\]
Kkutie7
  • Kkutie7
Yes I could have simplified it more. I also forgot to add the constant.
Kainui
  • Kainui
\[\int b^x dx = \frac{b^x}{\ln b}\] \[\int 3^xe^x dx = \int e^{(1+\ln 3)x}dx \] \[b=e^{1+\ln 3}\]

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