Find \({\displaystyle \int_0^2 \frac{5 x-3}{(x+1)(x-3)}\ dx}\). (4 marks)
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Find \({\displaystyle \int_0^2 \frac{5 x-3}{(x+1)(x-3)}\ dx}\). (4 marks)
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\(-\ln3 \)
\( \text{Let}\ \ \dfrac{5x-3}{(x+1)(x-3)} = \dfrac{A}{x+1}+\dfrac{B}{x-3}\ \ \ \text{for}\ \ A, B \in \mathbb{R} \)
\(\dfrac{5x-3}{(x+1)(x-3)}\) | \(=\dfrac{A}{x+1}+\dfrac{B}{x-3} \) | |
\(=\dfrac{A(x-3)+B(x+1)}{(x+1)(x-3)} \) |
\(\text{Equating numerators:}\)
\( A(x-3)+B(x+1)=5x-3 \)
\(\text{When}\ \ x=3, 4B=12\ \Rightarrow \ B=3 \)
\(\text{When}\ \ x=-1, -4A=-8\ \Rightarrow \ A=2 \)
\({\displaystyle \int_0^2 \frac{5 x-3}{(x+1)(x-3)}\ dx}\) | \( =\displaystyle \int_0^2 \dfrac{2}{x+1}+\dfrac{3}{x-3}\ dx \) | |
\(= \Big{[} 2\ln|x+1|+3\ln|x-3| \Big{]}_0^2 \) | ||
\(= 2\ln3+3\ln1-2\ln1-3\ln3 \) | ||
\(=-\ln3 \) |
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}`
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)`
`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)` |
Find `int 6/(x^2-9) dx`. (3 marks)
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`ln |\ (x-3)/(x + 3)\ |+ c`
`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` |
Evaluate `int_2^5 (x - 6)/(x^2 + 3x - 4)\ dx.` (4 marks)
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`2 ln (3/4)`
`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)` |
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)}`
`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.` |
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` |