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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 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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a.    `a = 3, b = 2`

b.    `3 log_e|x-1| + log_e |x^2+4|-1/2 tan^(-1)(x/2) +C`

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a.   `(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`  

 

b.     `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))`