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Calculus, SPEC1 2022 VCAA 4

Find `int(3x^(2)+4x+12)/(x(x^(2)+4))\ dx`.   (4 marks)

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`3ln |x|+2tan^(-1)((x)/(2))+c`

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`text{Using partial fractions:}`

`(3x^(2)+4x+12)/(x(x^(2)+4))` `-=(A)/(x)+(Bx+C)/(x^(2)+4)`  
`3x^(2)+4x+12` `=A(x^(2)+4)+x(Bx+C)`  
`3x^(2)+4x+12` `=(A+B)x^(2)+Cx+4A`  

 
`4A=12\ \ =>\ \ A=3`

`C=4`

`A+B=3\ \ =>\ \ B=0`

`:.\int(3x^(2)+4x+12)/(x(x^(2)+4))\ dx` `=int(3)/(x)+(4)/(4+x^(2))\ dx`  
  `=3ln |x|+2tan^(-1)((x)/(2))+c`  
Mean mark 55%.

Calculus, EXT2 C1 2022 HSC 12d

Using partial fractions, evaluate  `int_(2)^(n)(4+x)/((1-x)(4+x^(2))) dx`, giving your answer in the form  `(1)/(2)ln((f(n))/(8(n-1)^(2)))`, where `f(n)` is a function of `n`.  (4 marks)

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`1/2ln((4+n^2)/(8(1-n^2)))`

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`(4+x)/((1-x)(4+x^(2)))` `≡ A/(1-x) + (Bx+C)/(4+x^2)`  
`4+x` `≡A(4+x^2)+(Bx+C)(1-x)`  

 
`text{If}\ \ x=1, \ 5=5A\ \ =>\ \ A=1`

`(4+x)` `≡ 4+x^2+Bx-Bx^2+C-Cx`  
`4+x` `≡ (1-B)x^2+(B-C)x+C+4`  

 
`=>\ B=1, \ C=0`
 


Mean mark 85%.

`:.int_(2)^(n)(4+x)/((1-x)(4+x^(2))) dx`

`=int_2^n 1/(1-x) +x/(4+x^2)\ dx`

`=[-ln abs(1-x)+1/2ln(4+x^2)]_2^n`

`=-ln abs(1-n)+1/2ln(4+n^2)+lnabs(1-2)-1/2ln(4+2^2)`

`=-1/2ln(1-n)^2+1/2ln(4+n^2)-1/2ln(8)`

`=1/2ln((4+n^2)/(8(1-n^2)))`

Calculus, EXT2 C1 2021 HSC 11f

Express  `{3x^2-5}/{(x-2)(x^2 + x + 1)}`  as a sum of partial fractions over `RR`.  (3 marks)

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`{3x^2-5}/{(x-2)(x^2 + x + 1)} = {1}/{x-2} + {2x + 3}/{x^2 + x + 1}`

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`{3x^2-5}/{(x-2)(x^2 + x + 1)} = {A}/{(x-2)} + {B x + C}/{(x^2 + x + 1)}`

`A (x^2 + x + 1) + (Bx + C)(x-2) ≡ 3x^2-5`

`text{If} \ \ x = 2,`

`7A = 7 \ => \ A = 1`

`text{If} \ \ x = 0,`

`1-2C=-5 \ => \ C = 3`

`text(Equating coefficients:)`

`x^2 + x + 1 + Bx^2-2Bx + 3x -6 \ ≡ \ 3x^2 -5`

`(B + 1) x^2 + (4 -2B) x -5 \ ≡ \ 3x^2-5`

`=> B = 2`
 

`:. \ {3x^2-5}/{(x-2)(x^2 + x + 1)} = {1}/{x-2} + {2x + 3}/{x^2 + x + 1}`

Calculus, EXT2 C1 2003 HSC 1d

  1.  Find the real numbers  `a`  and  `b`  such that
     
    `qquad (5x^2-3x+13)/((x-1)(x^2+4)) ≡ a/(x-1) + (bx-1)/(x^2+4)`   (2 marks)

  2.  Hence find  `int (5x^2-3x+13)/((x-1)(x^2+4)) \ dx`   (2 marks)

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  1. `a = 3, b = 2`
  2. `3 log_e|x-1| + log_e |x^2+4|-1/2 tan^(-1)(x/2) +C`
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i.   `(5x^2-3x+13)/((x-1)(x^2+4)) ≡ (a(x^2+4) + (bx-1)(x-1))/((x-1)(x^2+4))`

 
`text(Equating numerators:)`

`5x^2-3x+13` `=ax^2+4a+bx^2-bx-x+1`  
  `=(a+b)x^2+(-b-1)x+4a+1`  

 

`-b-1` `=-3\ \ =>\ \ b=2`  
`a` `=3`  

 

ii.    `int (5x^2-3x+13)/((x-1)(x^2+4)) \ dx` `=int 3/(x-1)\ dx-int (2x)/(x^2+4)\ dx-int 1/(4+x^2)\ dx`
    `=3 log_e|x-1|-log_e |x^2+4|-1/2 tan^(-1)(x/2)+C`

Calculus, SPEC1 VCE SM-Bank 5

Find  `int(3x^2 + 8)/(x(x^2 +4))\ dx`.  (3 marks)

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` 2log_e x + 1/2log_e(x^2 + 4) + c` 

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`(3x^2 + 8)/(x(x^2 +4))` `=  a/x + (bx + c)/(x^2 + 4)`
`3x^2 +8` `=ax^2 + 4a+ bx^2 + cx`
  `=(a+b)x^2+cx+4a`
   

 `a + b = 3, \ c = 0, \ 4a = 8`

`:.a = 2, b = 1, c = 0`

`int(3x^2 + 8)/(x(x^2 +4))\ dx` `= int2/x\ dx + int x/(x^2 + 4)\ dx`
  `= 2log_e x + 1/2log_e(x^2 + 4) + c` 

Calculus, SPEC1 VCAA 2017 2

Find  `int_1^(sqrt3) 1/(x(1 + x^2))\ dx`, expressing your answer in the form  `log_e(sqrt(a/b))`  (4 marks)

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`ln(sqrt(3/2))`

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`text(Using partial fractions:)`

`int_1^(sqrt3) A/x + (Bx + C)/(1 + x^2)\ dx`

♦ Mean mark 48%.
MARKER’S COMMENT: A “large number” of students did not use partial fractions here.

`A(1 + x^2) + (Bx + C)x = 1`

`(A+B)x^2 + Cx+(A+C)=1`

`Cx=0\ \ =>\ C=0`

`A+C=1\ \ =>\ A=1`

`A+B=0\ \ =>\ B=-1`

 
`:. int_1^(sqrt3) 1/(x(1 + x^2))\ dx`

  `= int_1^(sqrt3) 1/x\ dx + int_1^(sqrt3) (−x)/(1 + x^2)\ dx`
  `= [ln|x|]_1^(sqrt3)-1/2 int_1^(sqrt3) (2x)/(1 + x^2)\ dx`
  `= ln(sqrt3)-ln(1)-1/2[ln(1 + x^2)]_1^(sqrt3)`
  `= ln(sqrt3)-1/2 ln(4) + 1/2 ln(2)`
  `= ln(sqrt3)-ln2 + ln(sqrt2)`
  `= ln((sqrt3 xx sqrt2)/2)`
  `= ln((sqrt6)/2)`
  `= ln(sqrt(3/2))`

Calculus, EXT2 C1 2010 HSC 1c

Find  `int 1/(x(x^2 + 1))\ dx`.   (3 marks)

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`ln\ |\ x\ | + 1/2ln\ (x^2 + 1) + c`

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`text(Using partial fractions:)`

`1/(x(x^2 + 1)) =` `a/x + (bx + c)/(x^2 + 1)`
`1=` `a(x^2+1)+x(bx+c)`
`1=` `(a + b)x^2 + cx + a`
`:.c = 0, \ \ a = 1,\ \ b = -1`

 

`:.int 1/(x(x^2 + 1))\ dx` `=int(1/x − x/(x^2 + 1))\ dx`
  `=ln\ |\ x\ |-1/2ln\ (x^2 + 1) + c`