smc-1055-10-Logs

Calculus, EXT2 C1 2020 HSC 11b

Use integration by parts to evaluate `int_1^e x ln x \ dx`. (3 marks)

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`frac{e^2 + 1}{4}`

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`u = ln \ x`	`v′ = x`
`u′ = frac{1}{x}`	`v = frac{x^2}{2}`

`int _1^e x \ ln \ x \ dx`	`= [ frac{x^2}{2} · ln \ x ]_1^e – int_1^e frac{x^2}{2} · frac{1}{x}\ dx`
	`= [frac{e^2}{2} ln \ e – frac{1}{2} ln 1]- int_1^e frac{x}{2}\ dx`
	`= frac{e^2}{2} – [ frac{x^2}{4}]_1^e`
	`= frac{e^2}{2} – ( frac{e^2}{4} – frac{1}{4} )`
	`= frac{e^2 + 1}{4}`

Calculus, EXT2 C1 2005 HSC 1c

Use integration by parts to evaluate `int_1^e x^7 log_e x dx`. (3 marks)

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`(7e^8 – 1)/(64)`

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	`u`	`= log_e x,`	`\ \ \ \ u^(′)`	`= (1)/(x)`
	`v^(′)`	`= x^7,`	`v`	`= (1)/(8) x^8`

`int_1^e x^7\ log_e x dx`	`= uv-int u^(′) v \ dx`
	`= [(x^8)/(8) ⋅ log_e x]_1 ^e-(1)/(8) int_1 ^e (1)/(x) ⋅ x^8 dx`
	`= ((e^8)/(8) log_e e-(1)/(8) log_e 1)-(1)/(8)[(1)/(8) x^8]_1 ^e`
	`= (e^8)/(8)-(1)/(8) ((e^8)/(8)-(1)/(8))`
	`= (e^8)/(8)-((e^8-1)/(64))`
	`= (7e^8 +1)/(64)`

Calculus, EXT2 C1 2003 HSC 1b

Use integration by parts to find `int x^3 log_e x dx` (3 marks)

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`(x^4 log_e x)/(4)-(x^4)/(16) + C`

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`u`	`= log_e x,`	`\ \ \ \ u^{′}`	`= (1)/(x)`
`v^{′}`	`= x^3,`	`v`	`= (x^4)/(4)`

`int x^3 log_e x dx`	`= uv-int u^{′}v \ dx`
	`= (x^4)/(4) log_e x-int (1)/(x) ⋅ (x^4)/(4) dx`
	`= (x^4 log_e x)/(4)-(1)/(4) int x^3 dx`
	`= (x^4 log_e x)/(4)-(x^4)/(16) + C`

Calculus, EXT2 C1 2011 HSC 1a

Find `int x ln x\ dx.` (2 marks)

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`(x^2 ln x)/2-x^2/4 + c`

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`text(Let)\ \ u`	`=lnx`	`v^{′}=`	`x`
`u^{′}`	`=1/x`	`v=`	`x^2/2`

`int x ln x\ dx`	`= x^2/2* ln x-int x^2/2 xx 1/x \ dx`
	`= (x^2 ln x)/2-1/2 int x\ dx`
	`= (x^2 ln x)/2-x^2/4 + c`

Calculus, EXT2 C1 2014 HSC 16c

Find `int(ln\ x)/((1 + ln\ x)^2)\ dx`. (3 marks)

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`x/(1+ln x) +c`

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`I=int(ln\ x)/((1 + ln\ x)^2)\ dx`

♦♦ Mean mark 31%.
STRATEGY: This is as challenging as integration gets … a substitution followed by applying integration by parts. Note that simply substituting `u=ln x` gets you a full mark.

`text(Let)\ \ \ u=`	`ln x`
`(du)/(dx)=`	`1/x=1/e^u`
`du=`	`1/e^u\ dx`
`dx=`	`e^u\ du`

`I`	`=int (u e^u)/(1+u)^2\ du`
	`=int ((1+u) e^u)/(1+u)^2\ du- int e^u/(1+u)^2\ du`
	`=int e^u/(1+u)\ du -int e^u/(1+u)^2\ du`

`text{Using integration by parts:}`

`u`	`=-e^u`	`u′`	`=-e^u`
`v′`	`=-(1+u)^-2`	`v`	`=(1+u)^-1`

`:.I`	`=int e^u/(1+u)\ du – ((-e^u)/(1+u) + int (e^u)/(1+u)\ du)`
	`=e^u/(1+u) +c`
	`=x/(1+ln x) +c`