int {frac{t}{e^{3t^2}}} , dt = | vedclass

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(D) Let $I = \int t e^{-3t^2} \, dt$. Substitute $z = -3t^2$,then $dz = -6t \, dt$,which implies $t \, dt = -\frac{1}{6} \, dz$. Substituting these in

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$-\frac{1}{6} e^{-3t^2} + c$

Vedclass · Answer

(D) Let $I = \int t e^{-3t^2} \, dt$. Substitute $z = -3t^2$,then $dz = -6t \, dt$,which implies $t \, dt = -\frac{1}{6} \, dz$. Substituting these into the integral: $I = \int e^z \left(-\frac{1}{6}\right) \, dz = -\frac{1}{6} \int e^z \, dz$. Integrating $e^z$ gives $e^z$: $I = -\frac{1}{6} e^z + c$. Substituting back $z = -3t^2$: $I = -\frac{1}{6} e^{-3t^2} + c$.

$\int {\frac{t}{e^{3t^2}}} \, dt = $

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