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sp0iltbrat

  • 3 years ago

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  1. sp0iltbrat
    • 3 years ago
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  2. across
    • 3 years ago
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    You're asking this in the wrong section, but here's your procedure anyway: Consider\[\int\frac{2+\ln u}{u^2}\,du.\]Let \(u=e^x\). Then, \(du=e^x\,dx\). Substituting this into the integral yields\[\int\frac{2+\ln e^x}{e^{2x}}e^x\,dx=\int\frac{2+x}{e^x}\,dx.\]

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