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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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)))`
`(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`
`:.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)))`
Express `{3x^2 - 5}/{(x - 2)(x^2 + x + 1)}` as a sum of partial fractions over `RR`. (3 marks)
`text{See Worked Solution}`
`{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}`
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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` |
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`
`(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` |
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)
`tan^-1 2 – tan^-1 1/2`
`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`
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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`
Find `int 1/(x(x^2 + 1))\ dx`. (3 marks)
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`ln\ |\ x\ | + 1/2ln\ (x^2 + 1) + c`
`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` |
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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` |