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Calculator question: Integral of (e^x^2-2x)/e^x^2 dx Answer is x+e^-x^2+C Please show work on how to get to the answer! Thank you
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Notice that \[ \frac{e^{x^2}-2 x}{e^{x^2}}=e^{-x^2} \left(e^{x^2}-2 x\right)=1-2 e^{-x^2} x \] Can you finish it now?
\[\Large \int \left(1-2 e^{-x^2} x\right) \, dx=e^{-x^2}+x +C \]
Okay so e^-x^2 times e^x^2 =1...how did you get the integral of 2e^-x^2 x?
If you take the integral of -2e^-x^2 x you would get -2e^-x^2 x^2/2 so the 2 will cancel but you would still have x^2 attached
Put \[ u=-x^2\\ du= -2 x dx\\ \int -2e^{-x^2} x dx =\int e^u du= e^u + C= e^{-x^2} + C \]
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