Can Veeky Forums integrate this?

From 0 to 1

Only intelligent people impress me.

Having an amazing memory makes understanding difficult concepts easier.

From 0 to 1, using 5 partial sums, evaluating at the midpoint of each partition and using only 4 decimal places of each sum I get 0.07616

It is pretty off but by hand I won't bother to get something better.

It is correct to 3 decimal places and from what I remember in high school physicists only use 2 decimal places so I guess you can call me the master integrator from now on.

wolframalpha.com/input/?i=Integral from 0 to 1 of x^2/(x^5 + 4)

You should be able to figure out the integral from this picture, unless of course you're not mathematically gifted.

>from what I remember in high school physicists only use 2 decimal places

What?

(1)/(5x^2)*(ln(x^5+4))

You're welcome

wolframalpha.com/input/?i=Derivative of 1/(5x^2)*(ln(x^5+4))

>You're welcome

What did he mean by this?

This is really just partial fractions. The bottom is quintic but has one obvious real root. You then factor the quartic and it will be ugly most likely but you seriously can just partial fractions this bitch

Back in high school for physics class they would tell us to just round to 2 decimal places most of the time.

(−√5−1)ln(∣∣x(2x+5√4(√5−1))+295∣∣)+(√5−1)ln(∣∣x(2x+5√4(−√5−1))+295∣∣)20⋅5√16+ln(∣∣x+225∣∣)5⋅5√16−(−8⋅√5−2135⋅5√4+8)arctan(4x−5√4(√5+1)2910√−√5+295⋅435−3)5⋅252√−√5+295⋅435−3⋅5√16+(−8⋅√5+2135⋅5√4−8)arctan(4x+5√4(√5−1)2910√√5+295⋅435−3)5⋅252√√5+295⋅435−3⋅5√16

ugly doesn't even begin to describe this

I see, just thought you meant that actual physicists only round to two decimal values

How can you guys claim that you're so smart yet you cant integrate that?

my bad

an amazing long term memory used to be valuable, but supply and demand bruh

now we have cheap memories that easily outclass humans

what matters for intelligence is that size of your working memory, which is completely different from the memory needed for reciting digits of pi

>my bad
You aren't wrong.

You are supposed to factorize the denominator and then partial fractions.

But lets just say, some fractions are more partial than others.

If you can't integrate it,then just leave Veeky Forums

I'm shit at this and not much of a Veeky Forums person in all honestly. I would Imagine you could do one of the following:

Rewrite x^2/x^5+4 as log|x^5+4|, then, when integrating turn it into an exponential e^x^5+4.

Alternatively multiply by bottom brackets and integrate the solution. I need help on this sorta stuff desu though so help would be appreciated.

this dumb weeaboo had been shitting up the board in search for attention for days now.
just letting you guys know so that you don't reply to him and he might just leave

This is my first time on the board in months, I've been doing further maths for while and this shit has been causing some grievance, sorry for the misunderstanding.

I was about to tell him how wrong he was but I see, he is just pretending to be retarded.

Or maybe he is retarded but he also knows that he is retarded so he is roleplaying as a retarded person who doesn't know he is retarded.

Really makes you think

No, I'm probably retarded, what am I doing wrong here? I've never been that great at maths, I just want a quick solution.

You can't
>Rewrite x^2/x^5+4 as log|x^5+4|

That doesn't even make sense.

Would it have to be in the context shown here then? (where you are integrating the exponent of an exponential).

Just residue theorem it nigga

What the fuck? That has nothing to do with this.

I figured that if you can integrate factors in that manner then you would be able to do the same in this case. Sure its something completely different, though I'd assume a link (though obviously false as proven).

Adding on to this, I'd assume the correct way is using partial fractions then integrating them individually?

In your example of differential equation that exponent of an integral is the general solution of that kind of DE.

You can't really put the log inside the integral and then put an e outside. It is not the same.

You could put an e and a log inside but then that would be completely useless.

Yes, this is what you have to do and there is no other way.

Great, thanks for the help! Sorry if I came off as retarded, I'm not very good at this.

Too lazy to compute the residues, but residue integration seems to work if you choose the contour [math]C_{R,\epsilon}[/math] that comprises a small semicircle around [math]-\sqrt[5](4)[/math] of radius e, then the segment [math][-\sqrt[5](4)+\epsilon, R][/math], then a semicircle from -R to R, then the segment [math][-R, -\sqrt[5](4)-\epsilon][/math].
As you let R go to infinity, the integral over the big semicircle goes to 0. Then you let epsilon go to 0 and then you have to compute the limit of the integral over the small semicircle (probably a dominated convergence argument). That tells you that the integral you want plus the limit of the small integral is 2pi i times the sum of the residues, which I won't compute but that wolfram can compute.
Still, it works very well.

Easy, just Taylor expand it and integrate term by term.

Take your pedophile cartoons back to

Yes. Any fractions of polynomials are easily solvable.

why can't i use integration by partial fractions here?

im a brainlet and i need an explanation

>all these retards who cant't do it
You literally just factor the denominator, decompose into partial fractions, separate the integral and integrate each part as u'/u or the inverse of a first or second degree polynomial which you should kill yourself if you're a math major and can't do it with your eyes closed.

holy shit im a complete brainlet and i got it holy shit

that's the solution btw

o well i lied

cant belive someone took the time to do that

...

let me believe what i want to believe

a computer took the time to do that
i don't use wolfram cause it's shit shows no steps
integral-calculator.com/#
just type it in there just to get a scope of how much work is required to get the solution