HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

leehuan · Apr 18, 2015

Re: MX2 2015 Integration Marathon

I was going to say 1 until I realised there was a vertical asymptote at x=3. How does that get addressed?

Drsoccerball · Apr 18, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Use the "plus 1 minus 1" trick to split the integrand and apply IBP once to get $$\frac{x}{1+\ln(x)}$ + $C$$

vat

Ekman · Apr 18, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
I was going to say 1 until I realised there was a vertical asymptote at x=3. How does that get addressed?

Fixed

leehuan · Apr 18, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
vat

$\int { \frac { \ln { x } }{ { \left( 1+\ln { x } \right) }^{ 2 } } dx }$
$=\int { \frac { 1+\ln { x } }{ { \left( 1+\ln { x } \right) }^{ 2 } } dx } -\int { \frac { 1 }{ { \left( 1+\ln { x } \right) }^{ 2 } } dx }$
$=\int { \frac { 1 }{ { 1+\ln { x } } } dx } -\int { \frac { 1 }{ { \left( 1+\ln { x } \right) }^{ 2 } } dx }$
$For\quad IBP\quad pick\quad u=\frac { 1 }{ { 1+\ln { x } } } ,\quad du=\frac { -1 }{ { x\left( 1+\ln { x } \right) }^{ 2 } } dx,\quad dv=dx,\quad v=x$
$=\frac { x }{ 1+\ln { x } } -\int { x\cdot \frac { -1 }{ { x\left( 1+\ln { x } \right) }^{ 2 } } dx } -\int { \frac { 1 }{ { \left( 1+\ln { x } \right) }^{ 2 } } dx } \quad$
$=\frac { x }{ 1+\ln { x } } +\int { \frac { 1 }{ { \left( 1+\ln { x } \right) }^{ 2 } } dx } -\int { \frac { 1 }{ { \left( 1+\ln { x } \right) }^{ 2 } } dx }$
$=\frac { x }{ 1+\ln { x } } +C\quad (statement\quad true\quad by\quad a\quad constant)$

RealiseNothing · Apr 18, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int{\frac{lnx}{(1+lnx)^2}}dx$

$\int \frac{1+lnx-\frac{x}{x}}{(1+lnx)^2}~dx$

$\frac{x}{1+lnx}+C$

Ekman · Apr 18, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int_{1}^{2} \frac{\sqrt{ln(9-x)}}{\sqrt{ln(9-x)}-\sqrt{ln(x+3)}}dx$

Bump

Drsoccerball · Apr 19, 2015

Re: MX2 2015 Integration Marathon

RealiseNothing said:
$\int \frac{1+lnx-\frac{x}{x}}{(1+lnx)^2}~dx$

$\frac{x}{1+lnx}+C$

reverse quotient rule :O ?

Sy123 · Apr 19, 2015

Re: MX2 2015 Integration Marathon

$\\ $Let$ \ I_n = \int_0^1 x^m (1-x)^n \ dx \\ \\ $Find a recurrence relation, where the relation is itself$ \\ $only in terms of$ \ n \ $(so no$ \ I(m,n))$

InteGrand · Apr 19, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ $Let$ \ I_n = \int_0^1 x^m (1-x)^n \ dx \\ \\ $Find a recurrence relation, where the relation is itself$ \\ $only in terms of$ \ n \ $(so no$ \ I(m,n))$

Are you just asking for a relation in terms of n and m (I don't think it's a "recurrence" relation if there's no earlier terms involved)? (In other words, you want the solution to the recurrence relation that can be formed?)

Sy123 · Apr 19, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Are you just asking for a relation in terms of n and m (I don't think it's a "recurrence" relation if there's no earlier terms involved)? (In other words, you want the solution to the recurrence relation that can be formed?)

The expression you get is in terms of m, and n, but the 'm' is a constant. The recurrence function should only be single variable

Ekman · Apr 19, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
The expression you get is in terms of m, and n, but the 'm' is a constant. The recurrence function should only be single variable

Is the answer:

$\frac{n!}{(n+m+1)!}$

Drsoccerball · Apr 19, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\\ $Let$ \ I_n = \int_0^1 x^m (1-x)^n \ dx \\ \\ $Find a recurrence relation, where the relation is itself$ \\ $only in terms of$ \ n \ $(so no$ \ I(m,n))$

The recurrence relation = $I_n= \frac{nI_{n-1}}{n+m+1}$
And you write an expresion for $I_{n-1}$ ect... to find the value desired

Drsoccerball · Apr 19, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Is the answer:

$\frac{n!}{(n+m+1)!}$

I got that

Drsoccerball · Apr 19, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int_{1}^{2} \frac{\sqrt{ln(9-x)}}{\sqrt{ln(9-x)}-\sqrt{ln(x+3)}}dx$

Bump is there a simple u substitution ?

Sy123 · Apr 19, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
The recurrence relation = $I_n= \frac{nI_{n-1}}{n+m+1}$
And you write an expresion for $I_{n-1}$ ect... to find the value desired

Yes you got it

Kaido · Apr 20, 2015

Re: MX2 2015 Integration Marathon

quick question, integral of tan^2x/(1+x^2)

Drsoccerball · Apr 20, 2015

Re: MX2 2015 Integration Marathon

Kaido said:
quick question, integral of tan^2x/(1+x^2)

Change tan^2x into sec^2x -1 and seperate the integral one the left side let u= tanx and the right side is just - tan inversex

leehuan · Apr 20, 2015

Re: MX2 2015 Integration Marathon

Kaido said:
quick question, integral of tan^2x/(1+x^2)

Drsoccerball said:
Change tan^2x into sec^2x -1 and seperate the integral one the left side let u= tanx and the right side is just - tan inversex

$\int { \frac { \tan ^{ 2 }{ x } }{ 1+{ x }^{ 2 } } dx } \\ =\int { \frac { \sec ^{ 2 }{ x } }{ 1+{ x }^{ 2 } } dx } -\int { \frac { dx }{ 1+{ x }^{ 2 } } } \\ u=\tan { x } \Rightarrow du=\sec ^{ 2 }{ x } dx\\ =\int { \frac { du }{ 1+{ \left( \tan ^{ -1 }{ u } \right) }^{ 2 } } } -\tan ^{ -1 }{ x }$

Wait, what?

Ekman · Apr 20, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int_{1}^{2} \frac{\sqrt{ln(9-x)}}{\sqrt{ln(9-x)}-\sqrt{ln(x+3)}}dx$

Bump

Kaido · Apr 20, 2015

Re: MX2 2015 Integration Marathon

Is it 1/2, took me a while but i think it looks good

Edit: I searched up some integration rules lol

HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

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