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Mathematics 8 Online
OpenStudy (anonymous):

int (1/x^2+x)dx

OpenStudy (anonymous):

you can writ it like this: int(x^-2)dx+int(x)dx=-1/x+(x^2)/2

hartnn (hartnn):

is it \(\large \int \dfrac{1}{x^2+x}dx\) ??

OpenStudy (anonymous):

really??? i think that its: int(1/x^2)+int(x) ???

hartnn (hartnn):

if its that, the you're correct :) else you'll need partial fractions...

OpenStudy (anonymous):

can you explain more about partial fraction?

hartnn (hartnn):

yes, for that we need to factor the denominator

hartnn (hartnn):

\(\large \dfrac{1}{x(x+1)}=\dfrac{A}{x}+\dfrac{B}{x+1}\) now you find A, B know how to ?

OpenStudy (anonymous):

yes Ax+A+Bx=1 so (A+B)x=0 A=1 so B=-1 1/x-1/x+1 ?

hartnn (hartnn):

absolutely correct! good work :) now those are easy to integrate.

OpenStudy (anonymous):

thanks hartnn

hartnn (hartnn):

welcome ^_^

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