Full solution

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

Simplify Integrands: Given the integral to solve: $\newline$ $I = \int_{50}^{\infty} 3t^{-4}(2+4t^{-3})^{-3} \, dt$ $\newline$ First, we need to simplify the integrand.
Further Simplify Expression: Simplify the integrand by rewriting $t^{-4}$ as $1/t^4$ and $t^{-3}$ as $1/t^3$ : $\newline$ $I = \int_{50}^{\infty} 3\left(\frac{1}{t^4}\right)\left(2+4\left(\frac{1}{t^3}\right)\right)^{-3} \, dt$
Make Substitution: Further simplify the expression inside the parentheses: $\newline$ $I = \int_{50}^{\infty} 3\left(\frac{1}{t^4}\right)\left(2+\frac{4}{t^3}\right)^{-3} \,dt$ $\newline$ = $\int_{50}^{\infty} 3\left(\frac{1}{t^4}\right)\left(2t^3+4\right)^{-3} \,dt$
Find Alternative Substitution: Now, let's make a substitution to simplify the integral. Let $u = 2t^3 + 4$ , then $du = 6t^2 dt$ . Since we have $t^4$ in the denominator, we need to express $t^2 dt$ in terms of $u$ to make the substitution.
Find Alternative Substitution: Now, let's make a substitution to simplify the integral. Let $u = 2t^3 + 4$ , then $du = 6t^2 dt$ . Since we have $t^4$ in the denominator, we need to express $t^2 dt$ in terms of $u$ to make the substitution.We have $t^2 dt$ in $du$ , but we need $t^4 dt$ in the integral. To get $t^4$ , we can express $t$ as $du = 6t^2 dt$ $0$ and then square it to get $du = 6t^2 dt$ $1$ . However, this approach is complex and might not be the best way to solve the integral. Let's try a different substitution.