e-x(4x-9)+C
e-x(4x-1)+C
We are given the integral \( \int (5 - 4x)e^{-x} \, dx \) and asked to find the solution.
We can solve this using the method of integration by parts. Let:
\( u = 5 - 4x \), so that \( du = -4 \, dx \),
and let \( dv = e^{-x} \, dx \), so that \( v = -e^{-x} \).
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:
\( \int (5 - 4x) e^{-x} \, dx = -(5 - 4x) e^{-x} - \int (-e^{-x})(-4) \, dx \).
Simplifying this, we get:
\( -(5 - 4x) e^{-x} + 4 \int e^{-x} \, dx \).
The integral of \( e^{-x} \) is \( -e^{-x} \), so we have:
\( -(5 - 4x) e^{-x} - 4e^{-x} + C \),
which simplifies to:
\( e^{-x}(4x - 9) + C \).
The correct answer is \( e^{-x}(4x - 9) + C \).