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Calculus, EXT2 C1 2023 HSC 13a

Find \({\displaystyle \int \frac{1-x}{\sqrt{5-4 x-x^2}}\ dx}\).  (3 marks)

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\(I=\sqrt{5-4x-x^2} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)

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\(I\) \(={\displaystyle \int \frac{1-x}{\sqrt{5-4 x-x^2}}\ dx}\)  
  \(= \dfrac{1}{2} {\displaystyle \int \frac{2-2x}{\sqrt{5-4 x-x^2}}\ dx}\)  
  \(=\dfrac{1}{2} {\displaystyle \int \frac{-4-2x}{\sqrt{5-4 x-x^2}}\ dx} + \dfrac{1}{2} {\displaystyle \int \frac{6}{\sqrt{5-4 x-x^2}}\ dx}\)  

 
\(\text{Let}\ \ u=5-4x-x^2 \)

\( \dfrac{du}{dx}=-4-2x\ \ \Rightarrow\ \ du=-4-2x\ dx \)

\(I\) \(= \dfrac{1}{2} {\displaystyle \int u^{-\frac{1}{2}}\ du} + {\displaystyle 3 \int \frac{1}{\sqrt{9-(x+2)^2}}\ dx}\)  
  \(= u^{\frac{1}{2}} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)  
  \(=  \sqrt{5-4x-x^2}+ 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)  

Calculus, EXT2 C1 2020 HSC 6 MC

Which expression is equal to `int frac{1}{x^2 + 4x + 10}\ dx`?

  1. `frac{1}{sqrt(6)} tan^-1 (frac{x + 2}{sqrt{6} )) + c`
  2. `tan^-1 (frac{x + 2}{sqrt{6} )) + c`
  3. `frac{1}{2 sqrt(6)} ln | frac{x + 2 - sqrt(6)}{x + 2 + sqrt(6)} | + c`
  4. `ln | frac{x + 2 - sqrt(6)}{x + 2 + sqrt(6)} | + c`
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`A`

Show Worked Solution
`int frac{1}{x^3 + 4x + 10}\ dx` `= int frac{1}{(x + 2)^2 + (sqrt6)^2}\ dx`
  `= frac{1}{6} tan^-1 (frac{x + 2}{sqrt6}) + c`

`=> \ A`