$\int \frac{3 t}{e^{4 t}} d t=-\frac{3}{4} t e^{-4 t}-\frac{3}{16} e^{-4 t}+C$ .

Name: $\int \frac{3 t}{e^{4 t}} d t=-\frac{3}{4} t e^{-4 t}-\frac{3}{16} e^{-4 t}+C$ .
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Question

Quizplus · Accepted Answer

This integral can be solved using integration by parts, where $u = 3t$ and $dv = e^{-4t} dt$. The solution correctly applies this method, resulting in $-\frac{3}{4} t e^{-4 t}-\frac{3}{16} e^{-4 t}+C$.

∫3te4tdt=−34te−4t−316e−4t+C\int \frac{3 t}{e^{4 t}} d t=-\frac{3}{4} t e^{-4 t}-\frac{3}{16} e^{-4 t}+C∫e4t3t​dt=−43​te−4t−163​e−4t+C

$\int \frac{3 t}{e^{4 t}} d t=-\frac{3}{4} t e^{-4 t}-\frac{3}{16} e^{-4 t}+C$