Hello /sci/! I've just now discovered my love for math,

>>7809578
we use the legendary plug and chug technique

someone should make a flow chart but in anycase:
1) observe that his is a fraction with a 4th degree polynomial in the denominator. therefore the flowchart say partial fraction that bitch
-(x + 1)/(x^2 + 1) + 1/x + x^(-2)

now we may integrate these 3 components seperately by plug-and-chugging them into our already known antiderivative patterns:
1/x => ln(x)
x^(-2) => -x^(-1)
-(x + 1)/(x^2 + 1) => -ln(x^2+1) + atan(x)

testingczs
[math]x = e^{x}[/math]

[math]\frac{x+1}{x^4 + x^2} = \frac{x}{x^4 + x^2} + \frac{1}{x^4 + x^2} = \frac{1}{x^2(x + 1)} + \frac{1}{x^2(x^2 + 1)} [/math]
Now comes the tricky part.
find A, B such that
[math]
\frac{1}{x^2(x + 1)} = \frac{A}{x^2} + \frac{B}{x+1}
[/math]
Then integrate.
Do the same for the second term.

>>7809578
Partial fraction decomposition

Thanks for the replies!
After allot of work i got this, can anyone tell me if i got it right?

File: solution antiderivative.png (2KB, 279x59px) Image search: [Google]

2KB, 279x59px

>>7809726
oops, forgot the image...

>>7809730
>missing the constant

>>7809739
That is also something i never quite understood, why add a constant? if you derivate it's going to disapear... This might be a really dumb question :/

>>7809748
Anon was joking; you got the joke.

>>7809748
Cause a family of functions like
2x+56 or 2x -9771
Hace the same derivate
So its the same with antiderivates

>>7809748
> if you derivate it's going to disapear
thats why we need it. why would you let the constant be zero, theres no reason for that

>>7809748
If you integrate, you can build the function on a cartesian graph, but you cant fully predict its position to the y axis.

When you derive, it goes away because you are describing other aspects of f(x) like its curvature and instantaneous rate of change.

>>7809748
in mathematics, you'rere usually looking for the most general answer there is. by computing the antiderivative, you're answering "find ALL functions with this derivative" rather than "find SOME function with this derivative". it can be proven that two differentiable functions have the same derivative if and only if they differ by a constant (it's a very easy proof and I suggest you try it yourself). therefore by adding a constant as a parameter, you're actually describing every function with the wanted derivative.

>>7809882
I see, thank you for the explanation!

>>7809959
Trying to search for general answer anywhere outside of mathematics is a lethal approach to the intelligence problem.

>>7809748
Initial/boundary conditions need to be satified.
Equations aren't there to just look pretty.

File: fking antiderivative.png (2KB, 48x44px) Image search: [Google]

2KB, 48x44px

Anyone to help me with this? Also an antiderivative.

>>7809654
There is a mistake in your maths. Don't skip steps anon, you aren't impressing anyone.

>>7810686
once again we use the legendary plug and chug technique

observe that there is a function and its derivative (if you cannot make these observations yet you have learn more plug-n-chug formulae).

we simply sub the function:
u=arctg(t)
du = 1/(1+t^2) dt

=>integrate u du = 0.5u^2 + k
=>sub t back in: 0.5arctg^2(t) + k

>>7809654
((x)(x+1))^-1 not ((x^2)(x+1))^-1
also, partial fractions don't work like that for the second term.