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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 EQ-Bank 1

Using partial fractions, show that

`int_0^(1/2) 8/(1-x^4)\ dx = log_e 9-4 tan^(−1) (1/2)`  (3 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`text(Using partial fractions:)`

`8/(1-x^4) = A/(1-x^2) + B/(1 + x^2)`

`A(1 + x^2) + B(1-x^2)` `= 8`
`A + B + (A-B)x^2` `= 8`
`A + B` ` = 8\ \ …\ (1)`
`A-B` ` = 0\ \ …\ (2)`

 
`A = 4, \ B = 4`

 

`4/(1-x^2) = A/(1-x) + B/(1 + x)`

`A(1 + x) + B(1-x)` `= 4`
`A + B + (A-B)x` `= 4`
`A + B` `= 4\ \ …\ (1)`
`A-B` `= 0\ \ …\ (2)`

 
`A = 2, B = 2`
 

`int_0^(1/2) 8/(1-x^4)\ dx` `= int_0^(1/2) 2/(1-x) + 2/(1 + x) + 4/(1 + x^2)\ dx`
  `= [−2ln |1-x| + 2ln |1 + x| + 4tan^(−1)x]_0^(1/2)`
  `= [2ln |(1 + x)/(1-x)| + 4tan^(−1)x]_0^(1/2)`
  `= 2ln |(1 1/2)/(1/2)| + 4tan^(−1)(1/2)-(2ln1 + 4tan^(−1) 0)`
  `= 2ln3 + 4tan^(−1)(1/2)`
  `= ln9 + 4tan^(−1)(1/2)`

Calculus, SPEC1 2011 VCAA 1

Find an antiderivative of  `(1 + x)/(9-x^2),\ \ x in R \ text(\{– 3, 3}).`  (3 marks)

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

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`int (1 + x)/(9-x^2)\ dx` `= int (1-x)/((3-x)(3 + x))\ dx`
  `= int A/(3-x) + B/(3 + x)\ dx`

 
`text(Using partial fractions:)`

`A(3 + x) + B(3-x) = 1 + x`

`text(When)\ \ x=3, \ 6A=4\ \ =>\ \ A=2/3`

`text(When)\ \ x=-3, \ 6B=-2\ \ =>\ \ B=-1/3`
 

`:. int (1 + x)/(9-x^2)\ dx`

`int 2/3 (1/(3-x))-1/3 (1/(3 + x))\ dx`

`= -2/3ln|3-x|-1/3ln|3 + x|,\ \ \ (c=0)`

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 2013 VCAA 2

Evaluate  `int_0^1 (x-5)/(x^2-5x + 6)\ dx.`  (4 marks)

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`ln(9/32)`

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`int_0^1 (x-5)/(x^2-5x + 6)\ dx` `= int_0^1 (x-5)/((x-3)(x-2))\ dx`
  `= int_0^1 A/(x-3) + B/(x-2)\ dx`

 
`text(Using partial fractions:)` 

`A(x-2) + B(x-3) = x-5`

`text(If)\ \ x = 2,` `-B` `= −3`
  `B` `= 3`
`text(If)\ \ x = 3,` `A` `=-2`

 
`:. int_0^1 (x-5)/(x^2-5x + 6)\ dx`

  `= int_0^1 −2/(x-3) + 3/(x-2)\ dx`
  `= [−2ln|x-3| + 3ln|x-2|]_0^1`
  `= −2ln(2) + 3ln(1) + 2ln(3)-3ln(2)`
  `= 2ln(3)-5ln(2)`
  `= ln(3^2/2^5)`
  `= ln(9/32)`

Calculus, SPEC1-NHT 2017 VCAA 2

Find  `a`  given that  `int_(-2)^a 8/(16-x^2)\ dx = log_e(6), \ a in (-2, 4)`.  (3 marks)

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 `a = 4/3`

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`I` `= int_(-2)^a 8/((4-x)(4 + x))\ dx`
  `= int_(-2)^a A/(4-x) + B/(4 + x)\ dx`

 
`text(Using partial fractions:)`

`A(4 + x) + B(4-x) = 8`

`text(If)\ \ x = 4\ \ =>\ \ 8A=8\ \ =>\ \ A=1`

`text(If)\ \ x = -4\ \ =>\ \ 8B=8\ \ =>\ \ B=1`
 

`:. I` `= int_(-2)^a 1/(4-x) + 1/(4 + x)\ dx`
`ln6` `= [ln |4 + x|-ln|4-x|]_(-2)^a`
  `= ln|4 + a|-ln|4-a|-(ln|2|-ln|6|)`
  `= ln ((4 + a)/(4-a))-ln (2/6),\ \ \ (a in (-2, 4))`
  `= ln ((3(4 + a))/(4-a))`
`6` `= 3((4 + a)/(4-a))`
`2` `=(4 + a)/(4-a)`
`8-2a` `=4+a`
`:. a` `=4/3`

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 2009 HSC 1d

Evaluate  `int_2^5 (x-6)/(x^2 + 3x-4)\ dx.`   (4 marks)

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`2 ln (3/4)`

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`text(Using partial fractions:)`

`(x-6)/(x^2 + 3x-4)` `=(x-6)/((x+4)(x-1))`
  `= A/(x+4) + B/(x-1)`
`:. x-6` `=A(x-1)+B(x+4)`

 

`text(When)\ \ x=1,\ \ 5B=-5\ =>B=-1`

`text(When)\ \ x=-4,\ \ -5A=-10\ =>A=2`

`:. int_2^5 (x-6)/(x^2 + 3x-4)\ dx` `= int_2^5 2/(x+4)\ dx-int_2^5 (dx)/(x-1)`
  `= [2 ln (x+4)]_2^5 -[ln(x-1)]_2^5` 
  `= 2(ln 9-ln 6)-(ln4-ln1)`
  `= 2ln(3/2)-ln4`
  `= 2ln(3/2)-2ln2`
  `= 2ln(3/4)`

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`

Show Worked Solution

`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`

Calculus, EXT2 C1 2011 HSC 7b

Let   `I = int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx.`

  1. Use the substitution  `u = 4-x`  to show that
     
          `I = int_1^3 (sin^2(pi/8 u))/(u(4-u))\ du.`  (2 marks)

     

  2. Hence, find the value of  `I`.  (3 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1/4 log_e 3`
Show Worked Solution

i.   `text(Let)\ \ u = 4-x,\ \ du = -dx,\ \ x = 4-u`

`text(When)\ \ x = 1,\ \ u = 3`

`text(When)\ \ x = 3,\ \ u = 1`

`I` `=int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx`
  `=int_3^1 (cos^2 (pi/8(4-u)))/((4-u)u) (-du)`
  `=-int_3^1 (cos^2(pi/2-pi/8 u))/((4-u)u)\ du`
  `=int_1^3 (sin^2(pi/8 u))/(u (4-u))\ du`

 

ii.   `text{S}text{ince}\ \ int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx = int_1^3 (sin^2(pi/8 x))/(x(4-x))\ dx`

 

`text(We can add the integrals such that)`

`2I` `=int_1^3 (cos^2(pi/8 x))/(x(4-x)) dx + int_1^3 (sin^2(pi/8 x))/(x(4-x)) dx`
  `=int_1^3 1/(x(4-x)) dx`

 

`text(Using partial fractions:)`

♦ Mean mark 36%.
`1/(x(4-x))` `=A/x+B/(4-x)`
`1` `=A(4-x)+Bx`

 
`text(When)\ \ x=0,\ \ A=1/4`

`text(When)\ \ x=0,\ \ B=1/4`

`2I` `=1/4 int_1^3 (1/x + 1/(4-x))\ dx`
  `=1/4 [log_e x-log_e (4-x)]_1^3`
  `=1/4 [log_e 3-log_e 1-(log_e 1-log_e 3)]`
  `=1/2 log_e 3`
`:.I` `=1/4 log_e 3`