# MX2 Integration Marathon 2021 (1 Viewer)

#### stupid_girl

##### Active Member
I made a typo in Wolfram earlier. It does give a nice answer.
$\bg_white \int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}$

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#### vernburn

##### Active Member
Not sure why Wolfram can't give a nice answer.
$\bg_white \int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}$
Let $\bg_white u = -x \implies du = -dx$:
$\bg_white I =\int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}dx$
$\bg_white =-\int_{1}^{-1} \frac{(-u-1)e^{-u}}{u^2+e^{-2u}}du$
$\bg_white =\int_{1}^{-1}\frac{(u+1)e^u}{e^{2u}u^2+1}du$
Notice that $\bg_white (u+1)e^u$ is the derivative of $\bg_white ue^u$ and hence the integral is in the form: $\bg_white \int \frac{f'(x)}{\left[f(x)\right]^2+1}dx = \tan^{-1}f(x)+C$. Thus,
$\bg_white I = \left[\tan^{-1}\left(ue^u\right)\right]_{1}^{-1}$
$\bg_white =-\tan^{-1}\left(\frac{1}{e}\right)-\tan^{-1}(e)$
$\bg_white =\boxed{-\frac{\pi}{2}}$ (using the well-known identity $\bg_white \tan^{-1}x+\tan^{-1}\frac{1}{x} = \frac{\pi}{2}$)

#### s97127

##### Member
Let $\bg_white u = -x \implies du = -dx$:
$\bg_white I =\int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}dx$
$\bg_white =-\int_{1}^{-1} \frac{(-u-1)e^{-u}}{u^2+e^{-2u}}du$
$\bg_white =\int_{1}^{-1}\frac{(u+1)e^u}{e^{2u}u^2+1}du$
Notice that $\bg_white (u+1)e^u$ is the derivative of $\bg_white ue^u$ and hence the integral is in the form: $\bg_white \int \frac{f'(x)}{\left[f(x)\right]^2+1}dx = \tan^{-1}f(x)+C$. Thus,
$\bg_white I = \left[\tan^{-1}\left(ue^u\right)\right]_{1}^{-1}$
$\bg_white =-\tan^{-1}\left(\frac{1}{e}\right)-\tan^{-1}(e)$
$\bg_white =\boxed{-\frac{\pi}{2}}$ (using the well-known identity $\bg_white \tan^{-1}x+\tan^{-1}\frac{1}{x} = \frac{\pi}{2}$)
how come a medical student like yourself still has so much interest in math after finishing HSC?

#### tito981

##### Active Member
how come a medical student like yourself still has so much interest in math after finishing HSC?
once u learn math properly u never forget.

#### CM_Tutor

##### Moderator
Moderator
I made a typo in Wolfram earlier. It does give a nice answer.
$\bg_white \int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}$
Interesting... integration calculator doesn't

#### idkkdi

##### Well-Known Member
how come a medical student like yourself still has so much interest in math after finishing HSC?
I have a feeling @vernburn likes maths more than the stuff he is going to do in med.

#### YonOra

##### Well-Known Member
Med is science, so how could u not get bored

#### Qeru

##### Well-Known Member
The ultimate tedious integral:

$\bg_white \int \sqrt[5]{\tan{x}} dx$

#### CoolKids101

##### New Member
The ultimate tedious integral:

$\bg_white \int \sqrt[5]{\tan{x}} dx$
You should attend the integration bee that's being held at UNSW lol

#### vernburn

##### Active Member
I have a feeling @vernburn likes maths more than the stuff he is going to do in med.
It does look like it doesn’t it! Jkjk
In reality, I really only like integration now because there is a certain elegance to it and skill required (and it’s still fresh in my head). I find the rest of maths quite boring and stale imho. My attempts at the above integrals are really just due to boredom (not long till uni starts now though).

The ultimate tedious integral:

$\bg_white \int \sqrt[5]{\tan{x}} dx$
I may be bored but not enough to attempt this monster!

#### Qeru

##### Well-Known Member
It does look like it doesn’t it! Jkjk
In reality, I really only like integration now because there is a certain elegance to it and skill required (and it’s still fresh in my head). I find the rest of maths quite boring and stale imho. My attempts at the above integrals are really just due to boredom (not long till uni starts now though).
Doesn't look like you were bored of maths considering you got a 99 for 4U. There had to be some interest there lol

#### CM_Tutor

##### Moderator
Moderator
The ultimate tedious integral:

$\bg_white \int \sqrt[5]{\tan{x}} dx$
This can easily be converted to a much more recognisable integral with the substitution:

\bg_white \begin{align*} \text{Let:} \qquad u^5 &= \tan^2{x} \\ \text{Implicitly differentiating with respect to x:} \qquad 5u^4\frac{du}{dx} &= 2\tan{x} \times \sec^2{x} \\ &= 2\sqrt{\tan^2{x}}(\tan^2{x} + 1) \\ &= 2\sqrt{u^5}(u^5+ 1) \\ \frac{5u^4}{2u^2\sqrt{u}(u^5 + 1)}du &= dx \\ dx &= \frac{5u^2}{2\sqrt{u}(u^5 + 1)}du \end{align*}

From which it follows that:

\bg_white \begin{align*} I &= \int{\sqrt[5]{\tan{x}}}\ dx \\ &= \int{\left(u^{\frac{5}{2}}\right)^{\frac{1}{5}} \times \frac{5u^2}{2\sqrt{u}(u^5 + 1)}}\ du \qquad \text{where u^5=\tan^2{x}} \\ &= \int{\sqrt{u} \times \frac{5u^2}{2\sqrt{u}(u^5 + 1)}}\ du \\ &= \frac{5}{2}\int{\frac{u^2}{u^5 + 1}}\ du \end{align*}

and the problem is reduced to a very (very) messy partial fractions problem. According to the integral calculator, and after cleaning up its awful formatting / presentation, the indefinite integral is:

$\bg_white I = \int{\sqrt[5]{\tan{x}}}\ dx$

$\bg_white = \frac{\sqrt{5}-1}{8}\ln\left[2\sqrt[5]{\tan^4{x}}-\left(\sqrt{5}-1\right)\sqrt[5]{\tan^2{x}}+2\right] + \frac{\sqrt{10+2\sqrt{5}}}{4}\tan^{-1}\left[\frac{4\sqrt[5]{\tan^2{x}}-\sqrt{5}-1}{\sqrt{10-2\sqrt{5}}}\right]$

$\bg_white - \left(\frac{\sqrt{5}+1}{8}\ln\left[2\sqrt[5]{\tan^4{x}}+\left(\sqrt{5}-1\right)\sqrt[5]{\tan^2{x}}+2\right] + \frac{\sqrt{10-2\sqrt{5}}}{4}\tan^{-1}\left[\frac{4\sqrt[5]{\tan^2{x}}+\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right]\right)$

$\bg_white + \frac{1}{2}\ln\left(\sqrt[5]{\tan^2{x}}+1\right)+C \qquad \text{for some constant C}$

#### stupid_girl

##### Active Member
I wrote this in the past but I have now forgotten how to solve it.
$\bg_white \int_{0}^{1}\log_{2}\left(\frac{\left(16^{\sqrt{x}}+k\cdot4^{\sqrt{x}}+4\right)\left(16^{1-\sqrt{x}}+k\cdot4^{1-\sqrt{x}}+4\right)}{\left(4^{\sqrt{x}}+k\cdot2^{\sqrt{x}}+4\right)\left(4^{1-\sqrt{x}}+k\cdot2^{1-\sqrt{x}}+4\right)}\right)dx=1$

Edit: I remember how to solve it now.

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#### CM_Tutor

##### Moderator
Moderator
I wrote this in the past but I have now forgotten how to solve it.
$\bg_white \int_{0}^{1}\log_{2}\left(\frac{\left(16^{\sqrt{x}}+k\cdot4^{\sqrt{x}}+4\right)\left(16^{1-\sqrt{x}}+k\cdot4^{1-\sqrt{x}}+4\right)}{\left(4^{\sqrt{x}}+k\cdot2^{\sqrt{x}}+4\right)\left(4^{1-\sqrt{x}}+k\cdot2^{1-\sqrt{x}}+4\right)}\right)dx=1$
Is the question here to prove this result - i.e. that the integral is 1 irrespective of the value of k - or is this an equation that needs to be solved to find the value(s) of k?

#### stupid_girl

##### Active Member
Is the question here to prove this result - i.e. that the integral is 1 irrespective of the value of k - or is this an equation that needs to be solved to find the value(s) of k?
The integral is 1 irrespective of the value of k.

#### s97127

##### Member
I wrote this in the past but I have now forgotten how to solve it.
$\bg_white \int_{0}^{1}\log_{2}\left(\frac{\left(16^{\sqrt{x}}+k\cdot4^{\sqrt{x}}+4\right)\left(16^{1-\sqrt{x}}+k\cdot4^{1-\sqrt{x}}+4\right)}{\left(4^{\sqrt{x}}+k\cdot2^{\sqrt{x}}+4\right)\left(4^{1-\sqrt{x}}+k\cdot2^{1-\sqrt{x}}+4\right)}\right)dx=1$

Edit: I remember how to solve it now.
can you post the solution here? This gives me a headache for the last few days

#### Qeru

##### Well-Known Member
can you post the solution here? This gives me a headache for the last few days
Yeah lol, some factoring trick I imagine, or King Rule but I dont see how that is used.

#### stupid_girl

##### Active Member
can you post the solution here? This gives me a headache for the last few days
What tricks have you tried?

The following techniques are NOT required.
integration by parts
differentiation under the integral sign
hyperbolic function

#### s97127

##### Member
What tricks have you tried?

The following techniques are NOT required.
integration by parts
differentiation under the integral sign
hyperbolic function
I've tried all the techniques above. I think the problem will be solved by substituting x with another variable t and we got I + I = 2 so I = 1.
Am i on the right path?

#### stupid_girl

##### Active Member
By adjusting the coefficient of k to 1, you should be able to take out log_2 (2), which is just 1. Integrating dx from 0 to 1 gives you 1.

The remaining factors have the same structure and you can offset them by suitable substitution.