How do I evaluate $\int (z + 1) e^{3 z} d z$ $int(z +1) e^(3z) dz$ ?

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How do I evaluate $\int (z + 1) e^{3 z} d z$ $int(z +1) e^(3z) dz$ ?

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Answer 1 · 2015-01-28T16:37:13

$\int (z + 1) e^{3 z} d z = \int z \cdot e^{3 z} d z + \int e^{3 z} d z$ $int (z+1)e^(3z) dz = int z*e^(3z)dz + int e^(3z)dz$

Second component is trivial
$\int e^{3 z} d z = \int \frac{e^{3 z}}{3} d (3 z) = \frac{e^{3 z}}{3}$ $int e^(3z) dz = int e^(3z)/3 d(3z) = e^(3z)/3$

For the first component, we use integration by part:
Recall: Integration by part: $\int u (d v) = u v - \int v (d u)$ $int u(dv) = uv - int v(du)$
let $u = z$ $u = z$ and $d v = e^{3 z}$ $dv = e^(3z)$ .
If $d v = e^{3 z}$ $dv = e^(3z)$ , then $v = \frac{e^{3 z}}{3}$ $v = e^(3z)/3$
(from second component)

Thus, $\int z \cdot e^{3 z} d z = z \cdot \frac{e^{3 z}}{3} - \int e^{3 z} d z$ $int z*e^(3z) dz = z*e^(3z)/3 - int e^(3z) dz$

Put the two component together:

$\int (z + 1) e^{3 z} d z$ $int (z+1)e^(3z) dz$
$= \int z \cdot e^{3 z} d z + \int e^{3 z} d z$ $= int z*e^(3z) dz + int e^(3z)dz$
$= [z \cdot \frac{e^{3 z}}{3} - \int e^{3 z} d z] + \int e^{3 z} d z$ $= [ z*e^(3z)/3 - int e^(3z) dz] + int e^(3z)dz$
$= z \cdot \frac{e^{3 z}}{3} + C$ $= z*e^(3z)/3 + C$

Hope this help!

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How do I evaluate $\int (z + 1) e^{3 z} d z$ $int(z +1) e^(3z) dz$ ?

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