Impossible question? (1 Viewer)

SammyT123 · May 18, 2016

$\int\frac{ln(x+2)}{x^2+4}$

found this on ekmans hard 4U qs compilation

P.S if someone has worked solutions to that pdf it would be a huge help

Edit: Questions are from here http://community.boredofstudies.org/14/mathematics-extension-2/345375/ekmans-compilation-mx2-questions.html

InteGrand · May 18, 2016

SammyT123 said:
$\int\frac{ln(x+2)}{x^2+4}$

found this on ekmans hard 4U qs compilation

P.S if someone has worked solutions to that pdf it would be a huge help

The original question had limits from 0 to 2. This makes it doable.

Without the limits, there's no elementary solution.

Tamama251 · May 18, 2016

I know the answer.
If x = infinity
Then answer = infinity

Ta da.

~ some idiot who doesn't even do 4 unit.

Paradoxica · May 18, 2016

SammyT123 said:
$\int\frac{ln(x+2)}{x^2+4}$

found this on ekmans hard 4U qs compilation

P.S if someone has worked solutions to that pdf it would be a huge help

Use the substitution

$u=\frac{4}{x}$

I'll show you the answer tomorrow at school if you don't get it.

si2136 · May 18, 2016

Tamama251 said:
I know the answer.
If x = infinity
Then answer = infinity

Ta da.

~ some idiot who doesn't even do 4 unit.

How do you integrate infinity lmao

@Integrand, when people say 0 to 2, the bottom one is 0 right?

Nailgun · May 18, 2016

si2136 said:
How do you integrate infinity lmao

@Integrand, when people say 0 to 2, the bottom one is 0 right?

yeah
well you can do it either way
but if you put the smaller number on top
it will come out multiplied by -1

Paradoxica · May 18, 2016

wait jks wrong type of integral

Carrotsticks · May 18, 2016

si2136 said:
How do you integrate infinity lmao

$\\ $Uhm obviously it's $ \infty x + C$

InteGrand · May 18, 2016

si2136 said:
How do you integrate infinity lmao

@Integrand, when people say 0 to 2, the bottom one is 0 right?

Yeah. But some people say "2 to 0" to mean the bottom one to be to 0. It should be said "0 to 2".

Paradoxica · May 18, 2016

Carrotsticks said:
$\\ $Uhm obviously it's $ \infty x + C$

Quite so, according to Wolfram Alpha.

si2136 · May 18, 2016

InteGrand said:
Yeah. But some people (mainly students learning this stuff) say "2 to 0" to mean the bottom one to be to 0. It should be said "0 to 2".

Yeah just confirming, because my teacher says the top to bottom and everyone follows it that way, but I was taught from the bottom

Nailgun · May 18, 2016

si2136 said:
Yeah just confirming, because my teacher says the top to bottom and everyone follows it that way, but I was taught from the bottom

yeah you were

(im sorry I couldn't resist)

InteGrand · May 18, 2016

si2136 said:
Yeah just confirming, because my teacher says the top to bottom and everyone follows it that way, but I was taught from the bottom

OK yeah, I guess quite a lot of people do say it that way.

But imagine a sum, people say like a sum from 1 to N (with 1 being the lower limit and N the upper limit). So similar thing with integrals I guess (integrals are basically continuous sums).

si2136 · May 18, 2016

Carrotsticks said:
$\\ $Uhm obviously it's $ \infty x + C$

$\\ $Amazing$ \displaystyle \int \infty ^ 2 = \infty x + C$

Where's the 2

[First time tex]

SammyT123 · May 18, 2016

Thanks for your replies everyone

I really love this forum because everyone helps out. I post a question and it gets answered so quickly.

Hoping one day I'll be able to give back (after learning Uni lvl maths lol)

Sent from my iPhone using Tapatalk

SammyT123 · May 18, 2016

InteGrand said:
The original question had limits from 0 to 2. This makes it doable.

Without the limits, there's no elementary solution.

How is this possible btw
(How can having limits make a qs doable )

So far I've only encountered questions where you pretty much solve the indefinite integral and then sub limits in to find the definite integral lol

And for anyone that cares I tried using IBP and got I=I
Rip HSC

InteGrand · May 18, 2016

SammyT123 said:
How is this possible btw
(How can having limits make a qs doable )

So far I've only encountered questions where you pretty much solve the indefinite integral and then sub limits in to find the definite integral lol

And for anyone that cares I tried using IBP and got I=I
Rip HSC

$\noindent The answer is based on using symmetry essentially (rather than finding an antiderivative). There's a lot of integrals that we don't have elementary functions for for the primitive, but can still evaluate the definite integral. In this one, let $I = \int_0 ^a \frac{\ln\left(x+a\right)}{x^2+a^2}\text{ d}x, a>0$. Then let $x=a\tan \theta$ for $\theta \in \left[0,\frac{\pi}{4}\right]$, so $\mathrm{d}x = a\sec^2 \theta \text{ d}\theta$, $x^2 + a^2 = a^2 \tan ^2 \theta + a^2 = a^2 \sec^2 \theta$. When $x=0$, $\theta = 0$ and when $x=a$, $\theta = \frac{\pi}{4}$. Hence$

$\begin{align*}I&= \int_{0} ^{\frac{\pi}{4}} \frac{\ln \left(a+a\tan \theta\right)}{a^2 \sec^2 \theta} \cdot a\sec^2 \theta \text{ d}\theta \\ &= \int _{0} ^{\frac{\pi}{4}} \frac{1}{a} \ln \left( a + a \tan \theta \right) \text{ d}\theta \\ &= \int _{0} ^{\frac{\pi}{4}} \frac{1}{a} \ln a \text{ d}\theta + \int _{0} ^{\frac{\pi}{4}} \frac{1}{a} \ln \left(1+\tan \theta\right) \text{ d}\theta \\ \Rightarrow I&= \frac{\pi}{4a}\ln a + \int _{0} ^{\frac{\pi}{4}} \frac{1}{a} \ln \left(1+\tan \theta\right) \text{ d}\theta.\end{align*}$

$\noindent Now we will use the property $\int _0 ^a f(x) \text{ d}x = \int _0 ^a f\left( a - x \right) \text{ d}x$. Note that (using compound angle formula for tan) $\ln \left(1+ \tan \left( \frac{\pi}{4} - \theta \right) \right) = \ln \left(1+\frac{1-\tan \theta}{1+\tan \theta} \right) = \ln \left( \frac{1+\tan \theta +1 - \tan \theta}{1+ \tan \theta} \right)= \ln 2 - \ln \left(1+\tan \theta \right)$. So$

$\begin{align*}J\equiv \int _{0} ^{\frac{\pi}{4}} \ln \left(1+\tan \theta\right) \text{ d}\theta &= \frac{\pi}{4}\ln 2 - J \\ \Rightarrow J &= \frac{\pi}{8}\ln 2. \end{align*}$

$\noindent So $I = \frac{\pi}{4a}\ln a + \frac{\pi}{8a}\ln 2$. The given integral is found by substituting $a=2$, to get an answer of $\frac{3\pi \ln 2}{16}$.$

SammyT123 · May 18, 2016

InteGrand said:
$\noindent The answer is based on using symmetry essentially (rather than finding an antiderivative). There's a lot of integrals that we don't have elementary functions for for the primitive, but can still evaluate the definite integral. In this one, let $I = \int_0 ^a \frac{\ln\left(x+a\right)}{x^2+a^2}\text{ d}x, a>0$. Then let $x=a\tan \theta$ for $\theta \in \left[0,\frac{\pi}{4}\right]$, so $\mathrm{d}x = a\sec^2 \theta \text{ d}\theta$, $x^2 + a^2 = a^2 \tan ^2 \theta + a^2 = a^2 \sec^2 \theta$. When $x=0$, $\theta = 0$ and when $x=a$, $\theta = \frac{\pi}{4}$. Hence$

$\begin{align*}I&= \int_{0} ^{\frac{\pi}{4}} \frac{\ln \left(a+a\tan \theta\right)}{a^2 \sec^2 \theta} \cdot a\sec^2 \theta \text{ d}\theta \\ &= \int _{0} ^{\frac{\pi}{4}} \frac{1}{a} \ln \left( a + a \tan \theta \right) \text{ d}\theta \\ &= \int _{0} ^{\frac{\pi}{4}} \frac{1}{a} \ln a \text{ d}\theta + \int _{0} ^{\frac{\pi}{4}} \frac{1}{a} \ln \left(1+\tan \theta\right) \text{ d}\theta \\ \Rightarrow I&= \frac{\pi}{4a}\ln a + \int _{0} ^{\frac{\pi}{4}} \frac{1}{a} \ln \left(1+\tan \theta\right) \text{ d}\theta.\end{align*}$

$\noindent Now we will use the property $\int _0 ^a f(x) \text{ d}x = \int _0 ^a f\left( a - x \right) \text{ d}x$. Note that (using compound angle formula for tan) $\ln \left(1+ \tan \left( \frac{\pi}{4} - \theta \right) \right) = \ln \left(1+\frac{1-\tan \theta}{1+\tan \theta} \right) = \ln \left( \frac{1+\tan \theta +1 - \tan \theta}{1+ \tan \theta} \right)= \ln 2 - \ln \left(1+\tan \theta \right)$. So$

$\begin{align*}J\equiv \int _{0} ^{\frac{\pi}{4}} \ln \left(1+\tan \theta\right) \text{ d}\theta &= \frac{\pi}{4}\ln 2 - J \\ \Rightarrow J &= \frac{\pi}{8}\ln 2. \end{align*}$

$\noindent So $I = \frac{\pi}{4a}\ln a + \frac{\pi}{8a}\ln 2$. The given integral is found by substituting $a=2$.$

Ah I see
Great explanation ,

Thanks once again

fluffchuck · May 18, 2016

the reason why you can't do it is because you forgot your dx sir

Paradoxica · May 18, 2016

My idea was to use a rational substitution to transform the integral, hence my (incorrect) substitution based on the scaled unit hyperbola.

Let us proceed with the generalised form of the integral.

$$\noindent$ \int_0^a \frac{\log{(x+a)}}{x^2+a^2}$d$x \\ y = \frac{a-x}{a+x} \equiv \frac{2a}{a+x}-1 \Leftrightarrow y+1 = \frac{2a}{a+x} \\ \frac{2a}{y+1} = a+x \Leftrightarrow x = a \,\frac{1-y}{1+y} \\ $d$x = -\frac{2a\,\text{d}y}{(1+y)^2} \\ $The borders run from 1 to 0, but the negative from the differential swaps the order.$ \\\int_0^a \frac{2a\log{\left(\frac{2a}{1+y} \right )}\text{d}y}{a^2\left( \left(\frac{1-y}{1+y} \right )^2+1 \right )(1+y)^2} = \frac{1}{a}\int_0^1 \frac{(\log{2a}-\log{(1+y))\text{d}y}}{y^2+1}$

$$\noindent$ x=ay \Rightarrow $ d$x = a\,$d$y \\ \int_0^1 \frac{\log{a(y+1)}}{a^2(y^2+1)}\,a\,$d$y = \frac{1}{a}\int_0^1 \frac{\log{a}-\log{(y+1)}}{y^2+1}\,$d$y$

Add the two forms of the integral together and the logarithms negate, leaving behind a convoluted constant on the numerator and y²+1 on the denominator. The integral is then trivial to evaluate, and the result follows.

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