When radium-226 decays to form radon-222, the radium nucleus emits a(an)

1 answer:

8 0

I believe the correct answer from the choices listed above is option A. When radium-226 decays to form radon-222, the radium nucleus emits an alpha particle. Alpha decay is a radioactive process in which a particle with two neutrons and two protons is ejected from the nucleus of a radioactive atom.

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How to solve this? \int \frac { 4 - 3 x ^ { 2 } } { ( 3 x ^ { 2 } + 4 ) ^ { 2 } } d x

ivanzaharov [21]

$\Large \mathbb{SOLUTION:}$

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