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Evaluate the integral using integration by parts where possible:

$\int(4x-1)e^{-x}\, dx$.

I have tried to solve this problem for hours but i still cant get the right answer. I would really appreciate it if someone could help explain how to do this step by step

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2 Answers 2

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HINT:

Use LIATE, $$\int (4x-1)e^{-x}\ dx=(4x-1)\int e^{-x}\ dx-\int\left(\frac{d(4x-1)}{dx}\cdot\int e^{-x}\ dx\right)\ dx=$$

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Let $u=4x-1$ and $dv=e^{-x}\,dx$. It follows, by the method of integration by parts, that we have $$\int (4x-1)e^{-x}\,dx =uv-\int v\,du\\=(4x-1)\int e^{-x}\,dx\,-\,4\int\left(\int e^{-x}\,dx\right)dx\\=-(4x-1)e^{-x}+4\int e^{-x}\,dx\\=(1-4x)e^{-x}-4e^{-x}+C\\=-(3+4x)e^{-x}+C,$$ for a constant $C$.

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