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Calculus, EXT2 C1 2025 HSC 12d

Find \(\displaystyle \int \frac{x^2-2 x+9}{(4-x)\left(x^2+1\right)} \, dx\).   (4 marks)

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\(2\, \tan ^{-1} x-\ln \abs{4-x}+C\)

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\(\dfrac{x^2-2 x+9}{(4-x)\left(x^2+1\right)}=\dfrac{A}{(4-x)}+\dfrac{B x+C}{x^2+1}\)

\(A\left(x^2+1\right)+(B x+C)(4-x)=x^2-2 x+9\)

\(\text{If}\ \  x=4:\)

\(17 A=16-8+9=17 \ \ \Rightarrow\ \ A=1\)

\(\text{If}\ \  x=0:\)

\(1+c \times 4=9 \ \ \Rightarrow\ \ C=2\)

\(\text{If}\ \  x=1:\)

\(2+3 B+6=8 \ \ \Rightarrow\ \ B=0\)

\(\displaystyle\int \frac{x^2-2 x+9}{(4-x)\left(x^2+1\right)}\, d x\) \(=\displaystyle\int \frac{1}{4-x}\, d x+\int \frac{2}{x^2+1}\, d x\)
  \(=2\, \tan ^{-1} x-\ln \abs{4-x}+C\)

Calculus, EXT2 C1 2024 HSC 12b

Use partial fractions to find \(\displaystyle \int \frac{3 x^2+2 x+1}{(x-1)\left(x^2+1\right)}\, d x\)   (3 marks)

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\(I=3 \ln \abs{x-1}+2 \tan ^{-1}(x)+c\)

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\(\displaystyle\int \dfrac{3 x^2+2 x+1}{(x-1)\left(x^2+1\right)}\, d x\)

\(\dfrac{3 x^2+2 x+1}{(x-1)\left(x^2+1\right)}=\dfrac{A}{(x-1)}+\dfrac{B x+C}{x^2+1}\)

  \(3 x^2+2 x+1\) \(=A\left(x^2+1\right)+(x-1)(B x+C)\)
    \(=A x^2+A+B x^2+C x-B x-C\)
    \(=(A+B) x^2+(C-B) x+A-C\)

 
\(\text {If } x=1:\)

\(3+2+1=2 A \ \Rightarrow \ A=3\)

\(A+B=3 \quad \quad \ \Rightarrow \ B=0\)

\(A-C=1 \quad \quad \ \Rightarrow \ C=2\)

  \(\therefore I\) \(=\displaystyle \int \frac{3}{x-1}+\frac{2}{x^2+1}\, d x\)
    \(=3 \ln \abs{x-1}+2 \tan ^{-1}(x)+c\)

Calculus, EXT2 C1 2022 SPEC1 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, EXT2 C1 2018 HSC 11c

By writing  `(x^2 - x - 6)/((x + 1)(x^2 - 3))`  in the form  `a/(x + 1) + (bx + c)/(x^2 - 3)`,

find  `int(x^2 - x - 6)/((x + 1)(x^2 - 3))\ dx`.  (4 marks)

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`2 ln |\ x + 1\ | – 1/2 ln |\ x^2 – 3\ | + c`

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`(x^2 – x – 6)/((x + 1)(x^2 – 3))` `≡ a/(x + 1) + (bx + c)/(x^2 – 3)`
`x^2 – x – 6` `≡ a(x^2 – 3) + (bx + c)(x + 1)`

 
`text(When)\ x = −1`

`1 + 1 – 6 = −2a\ \ =>\ \ a = 2`
 

`text(Equating co-efficients of)\ x^2`

`1 = (a + b)x^2\ \ =>\ \ b = −1`
 

`text(Equating co-efficients of)\ x`

`−1 = b + c\ \ =>\ \ c = 0`
 

`:. int(x^2 – x – 6)/((x + 1)(x^2 – 3))\ dx` `= int2/(x + 1) – x/(x^2 – 3)\ dx`
  `= 2 ln |\ x + 1\ | – 1/2 ln |\ x^2 – 3\ | + c`

Calculus, EXT2 C1 2007 HSC 1e

It can be shown that

`2/(x^3 + x^2 + x + 1) = 1/(x + 1) - x/(x^2 + 1) + 1/(x^2 + 1).`   (Do NOT prove this.)
 

Use this result to evaluate  `int_(1/2)^2 2/(x^3 + x^2 + x + 1)\ dx.`  (4 marks)

 

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`tan^-1 2 – tan^-1­ 1/2`

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`int_(1/2)^2 2/(x^3 + x^2 + x + 1)\ dx`

`=int_(1/2)^2 (1/(x + 1) – x/(x^2 + 1) + 1/(x^2 + 1)) dx`

`=[log_e(x + 1) – 1/2 log_e (x^2 + 1) + tan^-1 x]_(1/2)^2`

`=[(log_e 3 – 1/2 log_e 5 + tan^-1 2) – (log_e­ 3/2 – 1/2 log_e­ 5/4 + tan^-1­ 1/2)]`

`=log_e\ 3/sqrt5 -log_e (3/2 xx 2/sqrt5) + tan^-1 2 – tan^-1­ 1/2`

`=log_e (3/sqrt(5) xx sqrt (5)/3) + tan^-1 2 – tan^-1­ 1/2`

`=tan^-1 2 – tan^-1­ 1/2`

Calculus, EXT2 C1 2006 HSC 1c

  1. Given that  `(16x - 43)/((x - 3)^2 (x + 2))`  can be written as
     
    `qquad (16x - 43)/((x - 3)^2 (x + 2)) = a/(x - 3)^2 + b/(x - 3) + c/(x + 2)`,
     
    where  `a, b` and `c`  are real numbers, find  `a, b and c.`  (3 marks)

  2. Hence find  `int (16x - 43)/((x - 3)^2 (x + 2))\ dx.`  (2 marks)

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  1. `a = 1, b = 3, c = -3`
  2. `-1/(x-3)+3ln((x-3)/(x+2)) +c`
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i.   `(16x – 43)/((x – 3)^2 (x + 2)) = a/(x – 3)^2 + b/(x – 3) + c/(x + 2)`

`16x – 43 = a (x + 2) + b (x – 3) (x + 2) + c (x – 3)^2`
 

`text(When)\ \ x = 3,\ \ 5a =5\ \ =>a=1`

`text(When)\ \ x=–2,\ \ 25c=–75\ \ =>c=–3`

`text(When)\ \ x=0`

`-43` `= 2(1) – 6b + (-3)(-3)^2`
`6b` `= 18`
`b` `=3`

 
`:.a=1, b=3, c=–3`
 

ii.   `int (16x – 43)/((x – 3)^2 (x + 2))\ dx`

`=int (1/(x – 3)^2 + 3/(x – 3) – 3/(x + 2))\ dx`

`=-1/(x – 3) + 3ln(x – 3) -3ln(x + 2) + c`

`=-1/(x-3)+3ln((x-3)/(x+2)) +c`

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`

Calculus, EXT2 C1 2011 HSC 1c

  1. Find real numbers `a, b` and `c` such that 
      
       `1/(x^2 (x - 1)) = a/x + b/x^2 + c/(x - 1).`  (2 marks)

  2. Hence, find  `int 1/(x^2 (x - 1))\ dx`  (2 marks)

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  1. `a = −1,\ \ \ b = −1,\ \ \ c = 1`
  2. `log_e\ ((x – 1)/x) + 1/x + c`
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i.   `1/(x^2 (x – 1)) = a/x + b/x^2 + c/(x – 1)`

`1 = ax (x – 1) + b (x – 1) + cx^2`

`1 = ax^2 + cx^2 – ax + bx – b`

`1=(a+c)x^2+(b-a)x-b`
 

`text(Equating coefficients:)`

`=>0 = a + c,\ \ \ b – a = 0,\ \ \ -b = 1`

`:.a = -1,\ \ \ b = -1,\ \ \ c = 1`

 

ii.   `int 1/(x^2 (x – 1)) \ dx` `= int(-1/x – 1/x^2 + 1/(x – 1))\ dx`
  `= -log_e x + 1/x + log_e (x – 1) + c`
  `= log_e\ ((x – 1)/x) + 1/x + c`