Question

Solve:

Answer 1

\[\int\limits_{}^{}\frac{ 3 }{ 3+2x^2 }\]

Answer 2

do you know the standard result of
\(\huge \int \dfrac{1}{x^2+a^2}dx=...?\)
or if you don't want to use that you can do a trigonometric substitution to solve.

Answer 3

yes i know it, but 3 isn't a square number :P

Answer 4

and neither is 2. What substitution could i do??

Answer 5

You can always force things :D
\[\huge 3 = \sqrt3 ^2\]

Answer 6

\(\large 3+2x^2 = 2(\dfrac{3}{2}+x^2) =2[ (\sqrt {\dfrac{3}{2}})^2+x^2] \)
see what i did there ?

Answer 7

oh, i see. Well, that'll be painful to do, but since when is math not :P

Answer 8

math will not be painful anymore after you have enough practice ;)

Answer 9

I love math, but squaring and rooting at the same time is not cool man.

Answer 10

and @PeterPan, i'm loving the name ;)

Answer 11

I love it too :)

Answer 12

But sometimes, what seems to make things hard, actually makes things easier :D

Answer 13

true dat.

Answer 14

ok so now how do i solve \[\frac{ 1 }{ (x-3)^2 }\]

I know it's a really stupid question, but i just can't remember how to do it :P

Answer 15

FOIL?
Or actually, more specifically, square of the binomial.

Answer 16

you can substitute u= x-3

Answer 17

du=dx and i think you know how to integrate 1/u^2 du

Answer 18

omg i'm an idiot.