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

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

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

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

 
\(\text {If } x=1:\)

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

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

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

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

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 2023 HSC 11f

Find \({\displaystyle \int_0^2 \frac{5 x-3}{(x+1)(x-3)}\ dx}\).  (4 marks)

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\(-\ln3 \)

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\( \text{Let}\ \ \dfrac{5x-3}{(x+1)(x-3)} = \dfrac{A}{x+1}+\dfrac{B}{x-3}\ \ \ \text{for}\ \ A, B \in \mathbb{R} \)

\(\dfrac{5x-3}{(x+1)(x-3)}\) \(=\dfrac{A}{x+1}+\dfrac{B}{x-3} \)  
  \(=\dfrac{A(x-3)+B(x+1)}{(x+1)(x-3)} \)  

 
\(\text{Equating numerators:}\)

\( A(x-3)+B(x+1)=5x-3 \)

\(\text{When}\ \ x=3, 4B=12\ \Rightarrow \ B=3 \)

\(\text{When}\ \ x=-1,  -4A=-8\ \Rightarrow \ A=2 \)

\({\displaystyle \int_0^2 \frac{5 x-3}{(x+1)(x-3)}\ dx}\) \( =\displaystyle \int_0^2 \dfrac{2}{x+1}+\dfrac{3}{x-3}\ dx \)  
  \(= \Big{[} 2\ln|x+1|+3\ln|x-3| \Big{]}_0^2 \)  
  \(= 2\ln3+3\ln1-2\ln1-3\ln3 \)  
  \(=-\ln3 \)  

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)`

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`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, EXT2 C1 2019 HSC 11d

Find  `int 6/(x^2-9) dx`.  (3 marks)

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`ln |\ (x-3)/(x + 3)\ |+ c`

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`text(Using partial fractions):`

`6/(x^2-9)` `= 6/((x + 3)(x-3))`
  `= A/(x + 3) + B/(x-3)`

 
`:. A(x-3) + B(x + 3) = 6`
 

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

`text(When)\ \ x = -3,\ -6A = 6 \ => \ A = -1`

`int 6/(x^2-9)\ dx` `= int (-1)/(x + 3) + 1/(x-3)\ dx`
  `= -ln|\ x + 3\ | + ln |\ x-3\ | + C`
  `= ln |\ (x-3)/(x + 3)\ | + C`

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 5b

Show that  `int (dy)/(y(1 − y)) = ln (y/(1 − y)) + c`

for some constant  `c`, where  `0 < y < 1`.  (2 marks)

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`text{Proof (See Worked Solutions)}`

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`text(Solution 1)`

`d/(dy)[ln(y/(1 − y)) + c]` `= d/(dy)[ln y − ln(1 − y) + c]`
  `= 1/y − (-1)/(1 − y) + 0`
  `= (1 − y + y)/(y(1 − y))`
  `= 1/(y(1 − y))`

 

`text(Solution 2)`

`text(Using partial fractions:)`

`1/(y(1 − y))=` `A/y+B/(1-y)`
   

`=> A(y-1)+By=1`

`text(When)\ \ y=1,\ B=1`

`text(When)\ \ y=0,\ A=1`

`:.int (dy)/(y(1 − y))` `= int (1/y + 1/(1 − y))\ dy`
  `= ln y − ln(1 − y) + c`
  `= ln(y/(1 − y)) + c,\ \ \ \ \ y/(1 − y) > 0\ \ text(for)\ \ 0 < y < 1.`  

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 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`
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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`