Integral (1 Viewer)

[zeek]

Can someone please integrate this:

x^3 + x^2 + sin x
------------------------ dx
x^2 + 1

It has to be done over the real number field.... i tried putting it into mathematica and i got a crap result. Any help would be appreciated.

TIA

 

[Sober]

Yes, I just entered that into mathematica and got this:

I would not argue with that, the question does seem quite unconventional mixing polynomials with trig, where did you find this question?

 

[ianc]

What I did was a long division, to see if I could break that massive fraction down. (i always do long division if the degree of x is greater on the numerator than the denominator)

It ended up going down to:

x + 1 - x/(x^2+1) - 1/(x^2+1) + sinx/(x^2+1)

This is one hell of a question, because you're going to need to do integration by parts on sinx/(x^2+1) - and its one of those integration by parts where you have to do it twice, then substitute the second one back into the first....

To start it off, let u = sinx and dv/dx = 1/(x^2+1) (because you can integrate that nicely)

So it's still well within the scope of the extension 2 syllabus, but it would be worth quite a few marks in an exam.

 

[Shady01]

I tryd that u=sinx dv/dx= 1/x^2+1
du/dx= cosx v= tan^-1x

sinx.tan^-1x- Int[ tan^-1x.cosx] if u do that again can dv/dx be tan^-1x? thats where I get stuck?

 

[zeek]

yeah i had this in my exam and i've been killing myself trying to work it out... unfortunately it won't work :sad:

I've tried applying the product rule after breaking it up but it won't work as shady01 mentioned. I haven't tried the t-formulae because im guessing that would result in some kind of ugly transformation which will just be even more complicated. I haven't tried substituion because i haven't found a proper substitute. I still haven't tried splitting the numerator but im guessing that won't work because the denominator isn't trig based.

 

[Riviet]

I tryd that u=sinx dv/dx= 1/x^2+1
du/dx= cosx v= tan^-1x

sinx.tan^-1x- Int[ tan^-1x.cosx] if u do that again can dv/dx be tan^-1x? thats where I get stuck?

tan^-1x can easily be integrated by parts separately but even then it becomes much worse when you have expressions like x.sinx.tan^-1x and sinx.ln(1+x²) to integrate.

 

[zeek]

yup yup thats true... so does anyone have a solution or should i just stop worrying about it lol

 

[Shady01]

haha Yeah Riv your write. I always stuff Tan up becuase I screwed it up first attempt some months ago lol.