Question
Evaluate the following intregals:
$\int\frac{\text{x}+2}{\sqrt{\text{x}^2-1}}\text{dx}$
$\int\frac{\text{x}+2}{\sqrt{\text{x}^2-1}}\text{dx}$
✓
Answer
Let $\text{x}+2=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2-1)+\text {B}\ ...(1)$
$\Rightarrow\text{x}+2=\text{A}(2\text{x})+\text{B}$
Equating the coefficient of x and constant term on both sides, we obtain
$2\text{A}=1\Rightarrow\text{A}=\frac{1}{2}$
$\text{B}=2$
From (1), we obtain
$(\text{x}+2)=\frac{1}{2}(2\text{x})+2$
Then, $\int\frac{\text{x}+2}{\sqrt{\text{x}^2-1}}\text{dx}=\int\frac{\frac{1}{2}(2\text{x})+2}{\sqrt{\text{x}^2-1}}\text{dx}$
$=\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}+\int\frac{2}{\sqrt{\text{x}^2-1}}\text{dx}\ ...(2)$
$\text{In }\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}, $
Let $\text{x}^2-1=\text{t}\Rightarrow2\text{x}\text{ dx}=\text{dt}$
In $\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}$
let $\text{x}^2-1=\text{t}\Rightarrow2\text{x} \text{ dx}=\text{dt}$
$\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\frac{1}{2}\big[2\sqrt{\text{t}}\big]$
$=\sqrt{\text{t}}$
$=\sqrt{\text{x}^2-1}$
Then, $\int\frac{2}{\sqrt{\text{x}^2-1}}\text{dx}=2\int\frac{1}{\sqrt{\text{x}^2-1}}\text{dx}=2\log\big|\text{x}+\sqrt{\text{x }+1}\big|$
From equation (2), we obtain
$\int\frac{\text{x}+2}{\sqrt{\text{x}^2-1}}\text{dx}=\sqrt{\text{x}^2-1}+2\log\big|\text{x}+\sqrt{\text{x}^2-1}\big|+\text{C}$
$\Rightarrow\text{x}+2=\text{A}(2\text{x})+\text{B}$
Equating the coefficient of x and constant term on both sides, we obtain
$2\text{A}=1\Rightarrow\text{A}=\frac{1}{2}$
$\text{B}=2$
From (1), we obtain
$(\text{x}+2)=\frac{1}{2}(2\text{x})+2$
Then, $\int\frac{\text{x}+2}{\sqrt{\text{x}^2-1}}\text{dx}=\int\frac{\frac{1}{2}(2\text{x})+2}{\sqrt{\text{x}^2-1}}\text{dx}$
$=\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}+\int\frac{2}{\sqrt{\text{x}^2-1}}\text{dx}\ ...(2)$
$\text{In }\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}, $
Let $\text{x}^2-1=\text{t}\Rightarrow2\text{x}\text{ dx}=\text{dt}$
In $\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}$
let $\text{x}^2-1=\text{t}\Rightarrow2\text{x} \text{ dx}=\text{dt}$
$\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\frac{1}{2}\big[2\sqrt{\text{t}}\big]$
$=\sqrt{\text{t}}$
$=\sqrt{\text{x}^2-1}$
Then, $\int\frac{2}{\sqrt{\text{x}^2-1}}\text{dx}=2\int\frac{1}{\sqrt{\text{x}^2-1}}\text{dx}=2\log\big|\text{x}+\sqrt{\text{x }+1}\big|$
From equation (2), we obtain
$\int\frac{\text{x}+2}{\sqrt{\text{x}^2-1}}\text{dx}=\sqrt{\text{x}^2-1}+2\log\big|\text{x}+\sqrt{\text{x}^2-1}\big|+\text{C}$
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