smc-1056-30-PF given

Calculus, EXT2 C1 2008 HSC 1e

It can be shown that

`qquad (8(1-x))/((2-x^2)(2-2x+x^2)) = (4-2x)/(2-2x+x^2)-(2x)/(2-x^2)` (do NOT prove this)

Use this result to evaluate `int_0^1 (8(1-x))/((2-x^2)(2-2x+x^2))\ dx` (4 marks)

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`pi/2`

Show Worked Solution

`int_0^1 (8(1-x))/((2-x^2)(2-2x+x^2))\ dx`

`=int_0^1 (4-2x)/(2-2x+x^2)\ dx-int_0^1 (2x)/(2-x^2)\ dx`

`=int_0^1 (2-(2x-2))/(2-2x+x^2)\ dx + [log_e|2-x^2|]_0^1`

`=int_0^1 2/(1+(1-x)^2)-int_0^1 (2x-2)/(2-2x+x^2) + [log_e|2-x^2|]_0^1`

`=[-2tan^(-1)(1-x)]_0^1-[log_e(2-2x+x^2)]_0^1 + [log_e|2-x^2|]_0^1`

`=-2tan^(-1)0+2tan^(-1) 1-(log_e1-log_e2) + (log_e 1-log_e2)`

`=0+2 xx pi/4 + log_e2-log_e 2`

`=pi/2`

Calculus, EXT2 C1 2003 HSC 1d

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)

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Hence find `int (5x^2-3x+13)/((x-1)(x^2+4)) \ dx` (2 marks)

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`a = 3, b = 2`
`3 log_e|x-1| + log_e |x^2+4|-1/2 tan^(-1)(x/2) +C`

Show Worked Solution

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 2005 HSC 1b

Find real numbers `a` and `b` such that

`qquad (5x)/(x^2-x-6) ≡ a/(x-3) + b/(x+2)` (2 marks)

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Hence find `int (5x)/(x^2-x-6)\ dx` (1 mark)

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`a = 3, b = 2`
`3 log_e | x-3 | + 2 log_e | x+2 |+C`

Show Worked Solution

i. `(5x)/(x^2-x-6) ≡ (a(x+2) +b(x-3))/((x-3)(x+2))`

`text(Equating numerators:)`

`5x`	`=ax+2a+bx-3b`
`5x`	`=(a+b)x+2a-3b`

`a+b`	`=5\ \ …\ (1)`
`2a-3b`	`=0\ …\ (2)`

`(1) xx 2 – (2)`

`5b`	`=10`
`b`	`= 2`
`a`	`=3`

ii.	`int (5x)/(x^2-x-6)\ dx`	`=int 3/(x-3)\ dx + int 2/(x+2)\ dx`
		`=3 log_e \| x-3 \| + 2 log_e \| 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`

Show Worked Solution

`(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 2017 HSC 14a

It is given that `x^4 + 4 = (x^2 + 2x + 2) (x^2-2x + 2)`.

Find `A` and `B` so that `16/(x^4 + 4) = (A + 2x)/(x^2 + 2x + 2) + (B-2x)/(x^2-2x + 2)`. (1 mark)

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Hence, or otherwise, show that for any real number `m`

`int_0^m 16/(x^4 + 4)\ dx = ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`. (2 marks)

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Find the limiting value as `m -> oo` of

`int_0^m 16/(x^4 + 4)\ dx`. (1 mark)

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`A = 4 and B = 4`
`text(Proof)\ \ text{(See Worked Solutions)}`
`2 pi`

Show Worked Solution

i. `16/(x^4 + 4) = (A + 2x)/(x^2 + 2x + 2) + (B-2x)/(x^2-2x + 2)`

`text(Multiply each side by)\ \ x^4 + 4`

`16 = (A + 2x) (x^2-2x + 2) + (B-2x) (x^2 + 2x + 2)`

`text(When)\ \ x = 0,`

`2A + 2B`	`=16`
`A + B`	`= 8\ text{… (1)}`

`text(When)\ \ x = 1`

`(A + 2) (1) + (B-2) (5)`	`=16`
`A + 2 + 5B-10`	`=16`
`A + 5B`	`= 24\ text{… (2)}`

`text{Subtract (2) – (1)}`

`4B=16\ \ =>\ \ B=4`

`A=4`

ii. `int_0^m 16/(x^4 + 4)`

`= int_0^m (4 + 2x)/(x^2 + 2x + 2) dx + int_0^m (4-2x)/(x^2-2x + 2) dx`

`= int_0^m (2x + 2)/(x^2 + 2x + 2) + 2/(x^2 + 2x + 2) dx + int_0^m 2/(x^2-2x + 2)-(2x-2)/(x^2-2x + 2) dx`

`= [ln(x^2 + 2x + 2)]_0^m + int_0^m 2/(1 + (x + 1)^2) dx + int_0^m 2/(1 + (x-1)^2) dx-[ln (x^2-2x + 2)]_0^m`

`= [ln(m^2 + 2m + 2)-ln 2] + [2 tan^(-1) (x + 1)]_0^m + [2 tan^(-1) (x-1)]_0^m-[ln (m^2-2m + 2)-ln 2]`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 [tan^(-1) (m + 1)-tan^(-1) (1)] + 2 [tan^(-1) (m-1)-tan^(-1) (1)]`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1)-2 · pi/4 + 2 tan ^(-1) (m-1) + 2 · pi/4`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`

iii. `I = int_0^m 16/(x^4 + 4) = ln ((1 + 2/m + 2/m^2)/(1-2/m + 2/m^2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`

`lim_(m -> oo) I`	`= ln\ 1 + 2 · pi/2 + 2 · pi/2`
	`= 0 + pi + pi`
	`= 2 pi`

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`

Show Worked Solution

`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

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)

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

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

Show Worked Solution

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 2011 HSC 1c

Find real numbers `a, b` and `c` such that

`1/(x^2 (x - 1)) = a/x + b/x^2 + c/(x - 1).` (2 marks)

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

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

Show Worked Solution

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`