If int e^x left( frac{x^2-8x+19}{(x-1)^5} rig | vedclass

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(D) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$. Let $f(x) = \frac{x-4}{(x-1)^4}$. Then $f'(x) = \frac{(x-1)^4(1) - (x-4) \cdot 4(x-1)^3}

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(D) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$. Let $f(x) = \frac{x-4}{(x-1)^4}$. Then $f'(x) = \frac{(x-1)^4(1) - (x-4) \cdot 4(x-1)^3}{(x-1)^8} = \frac{(x-1) - 4(x-4)}{(x-1)^5} = \frac{x-1-4x+16}{(x-1)^5} = \frac{-3x+15}{(x-1)^5} = \frac{-3(x-5)}{(x-1)^5}$. Now, consider the integral $I = \int e^x \left( \frac{x^2-8x+19}{(x-1)^5} \right) dx$. We can rewrite the numerator as $x^2-8x+19 = (x-4)(x-1) - 3(x-5)$. So, $I = \int e^x \left( \frac{x-4}{(x-1)^4} - \frac{3(x-5)}{(x-1)^5} \right) dx = \int e^x [f(x) + f'(x)] dx = e^x f(x) + C = \frac{e^x(x-4)}{(x-1)^4} + C$. Comparing this with $\frac{e^x(lx+m)}{(x-1)^4} + C$, we get $l=1$ and $m=-4$. Therefore, $4l+m = 4(1) + (-4) = 4-4 = 0$.

If $\int e^x \left( \frac{x^2-8x+19}{(x-1)^5} \right) dx = \frac{e^x(lx+m)}{(x-1)^4} + C$, then $4l+m=$

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