Integrate. int3t^(-4)(2+4t^(-3))^(-3)

Identify Substitution: Given the integral to solve: \[ \int 3t^{-4}(2+4t^{-3})^{-3} \, dt \] Identify a substitution that can simplify the integral. Let $ u = 2+4t^{-3} $. Then, differentiate $ u $ with respect to $ t $ to find $ du $. \[ \frac{du}{dt} = -12t^{-4} \] \[ du = -12t^{-4} \, dt \]Find du: Solve for $ dt $ in terms of $ du $ and $ t $. \[ dt = \frac{du}{-12t^{-4}} \] Now, substitute $ u $ and $ dt $ into the integral. \[ \int 3t^{-4}(2+4t^{-3})^{-3} \, dt = \int 3t^{-4}u^{-3} \left( \frac{du}{-12t^{-4}} \right) \] \[ = \int \frac{-1}{4}u^{-3} \, du \]Substitute u and dt: Integrate $ \frac{-1}{4}u^{-3} $ with respect to $ u $. \[ \int \frac{-1}{4}u^{-3} \, du = \frac{-1}{4} \int u^{-3} \, du \] \[ = \frac{-1}{4} \left( \frac{u^{-2}}{-2} \right) + C \] \[ = \frac{1}{8}u^{-2} + C \]Integrate with u: Substitute back the original expression for $ u $ in terms of $ t $. \[ u = 2+4t^{-3} \] \[ \frac{1}{8}u^{-2} + C = \frac{1}{8}(2+4t^{-3})^{-2} + C \]

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Q. Integrate.

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\int 3 t^{-4}\left(2+4 t^{-3}\right)^{-3}

Identify Substitution: Given the integral to solve: $\newline$ $\int 3t^{-4}(2+4t^{-3})^{-3} \, dt$ $\newline$ Identify a substitution that can simplify the integral. $\newline$ Let $u = 2+4t^{-3}$ . Then, differentiate $u$ with respect to $t$ to find $du$ . $\newline$ $\frac{du}{dt} = -12t^{-4}$ $\newline$ $du = -12t^{-4} \, dt$
Find du: Solve for $dt$ in terms of $du$ and $t$ . $\newline$ $dt = \frac{du}{-12t^{-4}}$ $\newline$ Now, substitute $u$ and $dt$ into the integral. $\newline$ $\int 3t^{-4}(2+4t^{-3})^{-3} \, dt = \int 3t^{-4}u^{-3} \left( \frac{du}{-12t^{-4}} \right)$ $\newline$ $= \int \frac{-1}{4}u^{-3} \, du$
Substitute u and dt: Integrate $\frac{-1}{4}u^{-3}$ with respect to $u$ . $\newline$ $\int \frac{-1}{4}u^{-3} \, du = \frac{-1}{4} \int u^{-3} \, du$ $\newline$ $= \frac{-1}{4} \left( \frac{u^{-2}}{-2} \right) + C$ $\newline$ $= \frac{1}{8}u^{-2} + C$
Integrate with u: Substitute back the original expression for $u$ in terms of $t$ . $\newline$ $u = 2+4t^{-3}$ $\newline$ $\frac{1}{8}u^{-2} + C = \frac{1}{8}(2+4t^{-3})^{-2} + C$