Evaluating Indefinite integrals with primitive functions

Question

Evaluate: $$\int_{1}^{4} \frac{2}{\sqrt{(5x-4)^3}} dx$$

I don't even know where to go with this. I keep getting the primitive wrong I assume. Any help with working? I know I have to get to $\frac{3}{5}$ as a solution.

Answer 1

You're asking to solve

$$\int_{1}^{4} \frac{2}{(5x-4)^{3/2}} dx = \int_{1}^{4} 2(5x-4)^{-3/2} dx \tag{1}\label{eq1}$$

First, I would use substitution of $u = 5x - 4$ so $du = 5dx \implies dx = \frac{du}{5}$. Also, the limits will change to $1$ to $16$. Thus, \eqref{eq1} becomes

$$\begin{equation}\begin{aligned} \int_{1}^{16} \frac{2}{5}u^{-3/2} du & = \left(\frac{-4u^{-1/2}}{5}\right)\!\Big|_{1}^{16} \\ & = \left(\frac{-4}{5}\right)\left(16^{-1/2} - 1^{-1/2}\right) \\ & = \left(\frac{-4}{5}\right)\left(\frac{1}{4} - 1\right) \\ & = \frac{3}{5} \end{aligned}\end{equation}\tag{2}\label{eq2}$$

Note I used $\int u^{-3/2} du = -2u^{-1/2} + C$ in the above equation.