Question
Evaluate the following integral:
$\int\frac{1}{\sqrt{(2-\text{x})^2-1}}\text{ dx}$
$\int\frac{1}{\sqrt{(2-\text{x})^2-1}}\text{ dx}$
✓
Answer
$\int\frac{1}{\sqrt{(2-\text{x})^2-1}}\text{ dx}$
Let $2-\text{x}=\text{t}$
$-\text{dx}=\text{dt}$
$\text{dx}=-\text{dt}$
Now, $\int\frac{1}{\sqrt{(2-\text{x})^2-1}}\text{ dx}$
$=\int\frac{-\text{dt}}{\sqrt{\text{t}^2-1}}$
$=-\log\big|\text{t}+\sqrt{\text{t}^2-1}\big|+\text{C}$
$=-\log\big|(2-\text{x})+\sqrt{(2-\text{x})^2-1}\big|+\text{C}$
Let $2-\text{x}=\text{t}$
$-\text{dx}=\text{dt}$
$\text{dx}=-\text{dt}$
Now, $\int\frac{1}{\sqrt{(2-\text{x})^2-1}}\text{ dx}$
$=\int\frac{-\text{dt}}{\sqrt{\text{t}^2-1}}$
$=-\log\big|\text{t}+\sqrt{\text{t}^2-1}\big|+\text{C}$
$=-\log\big|(2-\text{x})+\sqrt{(2-\text{x})^2-1}\big|+\text{C}$
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