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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)
`{3x^2-5}/{(x-2)(x^2 + x + 1)} = {1}/{x-2} + {2x + 3}/{x^2 + x + 1}`
`{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}`
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` |