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Hi again,

so, I've been blasting away at the integration problems that I've been facing and just when I thought this term was going too easy, then I got to this one that kicked me in the left tooth.

Problem: \(\displaystyle \int_{0}^{1}4\ln(e^x)e^{-2x^2}dx\)

Attempted solution:

Ok, first off we move the 4 to the other side of the integral.

Now, I find \(\displaystyle u=e^{-2x^2}\frac{du}{dx}=\frac{-4x*\ln(e^{-2x^2})}{e^x}dx\)

\(\displaystyle =\frac{4}{-4}\int_{0}^{1}\frac{\ln(e^{-2x^2})}{e^x}\)

\(\displaystyle =-(\frac{\ln(e^{-2(1)^2})}{e^{(1)}}-\frac{\ln(e^{-2(0)^2})}{e^{(0)}})\)

Is this correct so far? It feels wrong, but I can't seem to find anywhere that specifically shows the integral of lne^x.

Thankd for any help...

so, I've been blasting away at the integration problems that I've been facing and just when I thought this term was going too easy, then I got to this one that kicked me in the left tooth.

Problem: \(\displaystyle \int_{0}^{1}4\ln(e^x)e^{-2x^2}dx\)

Attempted solution:

Ok, first off we move the 4 to the other side of the integral.

Now, I find \(\displaystyle u=e^{-2x^2}\frac{du}{dx}=\frac{-4x*\ln(e^{-2x^2})}{e^x}dx\)

\(\displaystyle =\frac{4}{-4}\int_{0}^{1}\frac{\ln(e^{-2x^2})}{e^x}\)

\(\displaystyle =-(\frac{\ln(e^{-2(1)^2})}{e^{(1)}}-\frac{\ln(e^{-2(0)^2})}{e^{(0)}})\)

Is this correct so far? It feels wrong, but I can't seem to find anywhere that specifically shows the integral of lne^x.

Thankd for any help...

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