HSC 2012-14 MX2 Integration Marathon (archive) (2 Viewers)

U MAD BRO · Aug 11, 2012

Re: MX2 Integration Marathon

Carrotsticks said:
Hey hey c'mon keep it friendly and elementary.

And for the record UMAD, you didn't properly show the acquiring of the Laurent Expension and the substituting of the poles to acquire the Residue.

Yes.

OK =)

U MAD BRO · Aug 11, 2012

Re: MX2 Integration Marathon

asianese said:
UMAD?!

I ain't even mad, are you?

Carrotsticks · Aug 11, 2012

Re: MX2 Integration Marathon

Both of you, stop.

Shadowdude · Aug 11, 2012

Re: MX2 Integration Marathon

To get back on track:

$\int \frac{dx}{1+x^{\frac{1}{4}}}$

gogogo

barbernator · Aug 11, 2012

Re: MX2 Integration Marathon

I will see you at the top of the state ranks for MX1 and MX2 umad

Carrotsticks · Aug 11, 2012

Re: MX2 Integration Marathon

Shadowdude said:
To get back on track:

$\int \frac{dx}{1+x^{\frac{1}{4}}}$

gogogo

Nice question.

Shadowdude · Aug 11, 2012

Re: MX2 Integration Marathon

Thanks! (i stole it from a reference book - i must admit... )

asianese · Aug 11, 2012

Re: MX2 Integration Marathon

$\int \frac{1}{1+x^{\frac{1}{4}}} dx \\ $Let u^4=x $ \therefore 4u^3 du = dx \\ I=4 \int \frac{u^3}{1+u} du\\ = 4\int u^2-u+1-\frac{1}{1+u} du $ by long division or synthetic division $ \\ = 4 [\frac{u^3}{3}-\frac{u^2}{2}+u-\ln (u+1)]du = 4[\frac{x^{\frac{3}{4}}}{3}-\frac{x^{\frac{1}{2}}}{2}+x^{\frac{1}{4}}-\ln ( x^{\frac{1}{4}}+1)]+C$

Next...(easier, but still good exercise)

$\int \log\sqrt{1+x^2} dx$

Shadowdude · Aug 11, 2012

Re: MX2 Integration Marathon

Yep!

Now:

$\int \frac{1}{\sqrt{e^{2x}-1}} dx$

Can be done in about five easy lines with a smart substitution.

Carrotsticks · Aug 11, 2012

Re: MX2 Integration Marathon

Shadowdude said:
Yep!

Now:

$\int \frac{1}{\sqrt{e^{2x}-1}} dx$

Can be done in about five easy lines with a smart substitution.

Haha I'm interested to see who gets it.

U MAD BRO · Aug 11, 2012

Re: MX2 Integration Marathon

let u=e^x

Shadowdude · Aug 11, 2012

Re: MX2 Integration Marathon

U MAD BRO said:
let u=e^x
and then a second sub making u^2-1=v

two substitutions?

simpler way.

asianese · Aug 11, 2012

Re: MX2 Integration Marathon

$\int \frac{dx}{\sqrt{e^{2x}-1}} \\ $Let u^2 = e^{2x}-1 $ \\ \therefore 2u du=2e^{2x}dx \\ udu = e^{2x}dx \\ $Now $u^2+1=e^{2x} $ from the substitution so$\\ I=\int \frac{udu}{1+u^2} \cdot \frac{1}{u} \\ = \int \frac{du}{1+u^2} \\= \tan^{-1}(\sqrt{e^{2x}-1})+C$

I really liked that question shadow!!!

U MAD BRO · Aug 11, 2012

Re: MX2 Integration Marathon

asianese said:
$\int \log\sqrt{1+x^2} dx$

I would use a hyperbolic function to make a substitution to get rid of the square root.

x=sinht
= xlnx-x+c

Carrotsticks · Aug 11, 2012

Re: MX2 Integration Marathon

You can get rid of the square root even faster.

(think of log laws)

asianese · Aug 11, 2012

Re: MX2 Integration Marathon

Lolwut...logs?

Carrotsticks · Aug 11, 2012

Re: MX2 Integration Marathon

asianese said:
Lolwut...logs?

$\log \sqrt{x} = \log x^{\frac{1}{2}} = \frac{1}{2} \log (x)$

asianese · Aug 11, 2012

Re: MX2 Integration Marathon

No I was saying that to UMAD and I didn't see your post carrot

U MAD BRO · Aug 11, 2012

Re: MX2 Integration Marathon

Try this one:

$\int{\frac{dx}{\sqrt{1 + x^4}}}$

barbernator · Aug 11, 2012

Re: MX2 Integration Marathon

U MAD BRO said:
Try this one:

$\int{\frac{dx}{\sqrt{1 + x^4}}}$

hmmm i can see a long way of doing that.

HSC 2012-14 MX2 Integration Marathon (archive) (2 Viewers)

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