evaluate the indefinite integral.\n int (6 - 7z)e^{-z} dz\n square + c

Explanation:

Step1: Apply integration - by - parts formula $\int u;dv=uv-\int v;du$

Let $u = 6 - 7z$ and $dv=e^{-z}dz$. Then $du=-7dz$ and $v=-e^{-z}$.

Step2: Substitute into the formula

$\int(6 - 7z)e^{-z}dz=(6 - 7z)(-e^{-z})-\int(-e^{-z})(-7dz)$.

Step3: Simplify the expression

$=-(6 - 7z)e^{-z}-7\int e^{-z}dz$.

Step4: Integrate $\int e^{-z}dz$

We know that $\int e^{-z}dz=-e^{-z}+C$. So $-(6 - 7z)e^{-z}-7(-e^{-z})+C$.

Step5: Expand and combine like - terms

$=-6e^{-z}+7ze^{-z}+7e^{-z}+C=(7z + 1)e^{-z}+C$.

Answer:

$(7z + 1)e^{-z}$