Mathematicians here ? I'm having a real trouble with this

>>8940378
try more degrees of freedom with the first fraction like [math] \frac{at+b}{(1+t)^2} [/math]

if that doesn't work just look up method of partial fractions

>>8940382
ok what I did is to type : [math](1+t)^2(1+t^2)[/math] as

[math]t^4+2t^3+2t^2+1[/math]

and then I divided it by : [math]t^2+2t[/math] in order to find another way of writing it.

So in the end I found :

[math](1+t)^2(1+t^2)[/math] = [math](t^2+2t)(t^2+2)+1[/math]

and now I replace [math]t^2[/math] with [math]X[/math]

>>8940389
am I going the right way?

>>8940389
Wait wait, I fucked it up

it should be :

[math](1+t)^2(1+t^2)[/math] = [math](t^2+2t)(t^2+2-\frac {2}{t})+5[/math]

I'm fucking lost

>>8940396
ok so i did the method of partial fractions (https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html) with two degrees of freedom ([math] \frac{a}{(1+t)^2} + \frac{b}{(1+t^2)} [/math]) and then with three degrees of freedom ([math] \frac{at+b}{(1+t)^2} + \frac{c}{(1+t^2)} [/math]) and found you don't have enough degrees of freedom. i didn't notice the singular [math] t^3 [/math] dependence with three degrees of freedom so wasted time doing it. anyway the decomposition requires 4 variables to solve for so it's not going to be on your test. an easier version may be in which case you should notice to solve for it with this method it's always the degree-1 polynomial in the numerator when compared to the polynomial in the denominator (pretty sure anyway)

>>8940442
this was in a test last year and professor said it will be almost same this year

>>8940445
kek really?

well hopefully it's not with partial fractions then

i checked my algebra just to be sure and it can't be two because it doesn't work and can't be three because one variable simplifies to 0 and then the two degrees case so it has to be at least 4

>>8940468
well the quetion says "If possible"

>>8940378
[eqn]\int_0^\infty \frac{1}{(1+t)^2(1+t^2)} dt [/eqn]

[eqn] \int_0^\infty \frac{-t}{2(t^2+1)} dt + \int_0^\infty \frac{1}{2(t+1)} dt + \int_0^\infty \frac{1}{2(t+1)^2} dt [/eqn]

[eqn] -\frac{1}{4} \ln(t^2+1)+\frac{1}{2} \ln(t+1)-\frac{1}{2(t+1)} |_{t=0}^{\infty} [/eqn]

[eqn] -\frac{1}{2}+\lim_{t \rightarrow \infty} \ln \left(\frac{\sqrt{t+1}}{\sqrt[4]{t^2+1}} \right) [/eqn]

I'm too lazy to do the last part, but it's just L'Hospital's Rule.

>>8940470
to use method of partial fractions?
damn

anyhow i'll show you the logic then

assume 2 degrees

[eqn] \frac{a}{(1+t)^2} + \frac{b}{(1+t^2)} = \frac{1}{(1+t)^2(1+t^2)} [/eqn]

multiplying to get common denominator on left side and simplifying

[eqn] \implies at^2+bt^2+2bt+a+b = 1 [/eqn]

with the requirement [math] t \neq 1 [/math] but that's fine since it's only defined from 0 to infinity

you see from the [math] t [/math] term b has to equal 0 which implies a has to equal 0 so the whole thing doesn't work

repeating the same thing for 3 degrees [math] \frac{at+b}{(1+t)^2} + \frac{c}{(1+t^2)} [/math]

yields the same thing for the [math] t^3 [/math] term which sets a=0 and therefore simplifies to the 2 degree case and therefore still doesn't work

so you have to have at least 4 degrees of freedom to even try to solve it which implies 4 variables and 4 equations needed to solve

>>8940483
oh shit i didn't think about splitting like that

that method probs works for partial fractions

>>8940486
>>8940483

Thanks a lot guys, do you have any tips on how to become as good as you in things like these ?

>>8940495
practice is the best way for calc 2

i didn't especially like calc 2 because of how computational and that integral is an example why. unless you enjoy how computational it was, these types of integrals stop for the most part so you don't need to sweat being super good at integrating.

>>8940509
alright, I'll try my best to be better, thank you