Evaluate the integral. int3xe^(-4x)dx Answer:

Choose u and dv: Let's use integration by parts to solve the integral of 3xe^(-4x)dx. Integration by parts is given by the formula ∫u dv = uv - ∫v du, where u and dv are parts of the integrand that we choose. We will let u = 3x, which means du = 3dx, and dv = e^(-4x)dx, which means v = -1/4 e^(-4x) after integrating dv.Apply integration by parts: Now we apply the integration by parts formula. We have: ∫3xe^(-4x)dx = uv - ∫v du = (3x)(-1/4 e^(-4x)) - ∫(-1/4 e^(-4x))(3dx)Simplify the integral: Simplify the integral: ∫3xe^(-4x)dx = -3/4 x e^(-4x) - ∫(-3/4 e^(-4x))dxIntegrate the remaining term: Now we integrate the remaining term -3/4 e^(-4x) with respect to x: ∫(-3/4 e^(-4x))dx = (-3/4) * (-1/4) * e^(-4x) = 3/16 e^(-4x)Combine the two parts: Combine the two parts to get the final answer: ∫3xe^(-4x)dx = -3/4 x e^(-4x) + 3/16 e^(-4x) + C, where C is the constant of integration.

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Evaluate the integral.

int3xe^(-4x)dx
Answer:

Evaluate the integral. $\newline$ $\int 3 x e^{-4 x} d x$ $\newline$ Answer:

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Calculus

Find indefinite integrals using the substitution and by parts

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Q. Evaluate the integral.

\newline

\int 3 x e^{-4 x} d x

\newline

Answer:

Choose $u$ and $dv$ : Let's use integration by parts to solve the integral of $3xe^{-4x}\,dx$ . Integration by parts is given by the formula $\int u\, dv = uv - \int v\, du$ , where $u$ and $dv$ are parts of the integrand that we choose. We will let $u = 3x$ , which means $du = 3\,dx$ , and $dv = e^{-4x}\,dx$ , which means $v = -\frac{1}{4} e^{-4x}$ after integrating $dv$ .
Apply integration by parts: Now we apply the integration by parts formula. We have: $\newline$ $\int 3xe^{-4x}\,dx = uv - \int v\,du$ $\newline$ = $(3x)(-\frac{1}{4} e^{-4x}) - \int(-\frac{1}{4} e^{-4x})(3\,dx)$
Simplify the integral: Simplify the integral: $\int 3xe^{-4x}dx = -\frac{3}{4} x e^{-4x} - \int(-\frac{3}{4} e^{-4x})dx$
Integrate the remaining term: Now we integrate the remaining term $-\frac{3}{4} e^{-4x}$ with respect to $x$ : $\int\left(-\frac{3}{4} e^{-4x}\right)dx = \left(-\frac{3}{4}\right) * \left(-\frac{1}{4}\right) * e^{-4x} = \frac{3}{16} e^{-4x}$
Combine the two parts: Combine the two parts to get the final answer: $\int 3xe^{-4x}dx = -\frac{3}{4} x e^{-4x} + \frac{3}{16} e^{-4x} + C$ , where $C$ is the constant of integration.