# Question #64888

Mar 22, 2017

$\frac{1}{\left(1 + x\right) \left(3 + x\right)} = \frac{1}{2 \left(1 + x\right)} - \frac{1}{2 \left(3 + x\right)}$

$y = 2 \sqrt{2} \sqrt{\frac{1 + x}{3 + x}}$

#### Explanation:

$\frac{1}{\left(1 + x\right) \left(3 + x\right)} = \frac{\alpha}{1 + x} + \frac{\beta}{3 + x}$

$= \frac{\alpha \left(3 + x\right) + \beta \left(1 + x\right)}{\left(1 + x\right) \left(3 + x\right)} = \frac{1}{\left(1 + x\right) \left(3 + x\right)}$

Look at numerators and let $x = - 1$

$\implies 2 \alpha = 1 , \alpha = \frac{1}{2}$

Look at numerators and let $x = - 3$

$\implies - 2 \beta = 1 , \beta = - \frac{1}{2}$

$\implies \frac{1}{\left(1 + x\right) \left(3 + x\right)} = \frac{1}{2 \left(1 + x\right)} - \frac{1}{2 \left(3 + x\right)}$

$y ' = y \left(\frac{1}{2 \left(1 + x\right)} - \frac{1}{2 \left(3 + x\right)}\right)$

This separates as:

$\frac{2 y '}{y} = \frac{1}{\left(1 + x\right)} - \frac{1}{\left(3 + x\right)}$

We integrate wrt x:

$2 \int \frac{1}{y} y ' \mathrm{dx} = \int \frac{1}{\left(1 + x\right)} - \frac{1}{\left(3 + x\right)} \mathrm{dx}$

By chain rule:

$= 2 \int \frac{1}{y} \mathrm{dy} = \int \frac{1}{\left(1 + x\right)} - \frac{1}{\left(3 + x\right)} \mathrm{dx}$

$\implies 2 \ln y = \ln \left(1 + x\right) - \ln \left(3 + x\right) + C$

With $C$ as a generic constant, ie the actual letter is oblivious to operations:

$\implies \ln y = \ln \sqrt{\frac{1 + x}{3 + x}} + C$

Put both sides to power of $e$:

${e}^{\ln y} = {e}^{\ln \sqrt{\frac{1 + x}{3 + x}} + C}$

$y = {e}^{\ln \sqrt{\frac{1 + x}{3 + x}}} {e}^{C} = C \sqrt{\frac{1 + x}{3 + x}}$

$y \left(1\right) = 2 \implies 2 = C \cdot \sqrt{\frac{2}{4}} \implies C = 2 \sqrt{2}$

$\implies y = 2 \sqrt{2} \sqrt{\frac{1 + x}{3 + x}}$