$\begin{array}{l} \displaystyle \int \dfrac{4 - 3x^2}{(3x^2 + 4)^2} dx \\ \\ = \displaystyle \int \dfrac{4 - 3x^2}{x^2\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ = \displaystyle \int \dfrac{\dfrac{4}{x^2} - 3}{\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ \text{Let }u = 3x + \dfrac{4}{x} \implies du = \left(3 - \dfrac{4}{x^2}\right)\ dx \\ \\ \text{So the integral becomes} \\ \\ = \displaystyle -\int \dfrac{du}{u^2} \\ \\ = -\dfrac{u^{-2 + 1}}{-2 + 1} + C \\ \\ = \dfrac{1}{u} + C \\ \\ = \dfrac{1}{3x + \dfrac{4}{x}} + C \\ \\ = \boxed{\dfrac{x}{3x^2 + 4} + C}\end{array}$

5 0

3 years ago

NEED HELP! ASAP! 50 POINTS!

Lerok [7]

Answer:

$y=13\text{ miles per gallon}$

Step-by-step explanation:

We have the two points: (0, 11) and (3, 50).

And we want to find the rate of change in y relative to x, where x is the gallons and y is the miles.

In simple terms, we want to find the slope between the two points.

So, let's use the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

Let (0, 11) be (x₁, y₁) and let (3, 50) be (x₂, y₂). Substitute. Let's also put "miles" in the numerator with the Ys and "gallons" in the denominator with the Xs. This yields:

$y=\frac{50-11\text{ miles}}{3-0\text{ gallons}}$