1u c...
x2+2x+3=(x+1)2+2
r8?
x+1=t
∫t4/(t2+2)3 dt
then wat???
14 Answers
1no i tried in that tanθ way earlier....
getting,,,
∫sin4xdx at last....
simple but looooooooong...
i want sumthing shorter...
can u gv ny other tricky way????
62how about partial fractions? i think they could work...
I havent written down...
Try using the sum of
1/t-4/t2+4/t3
where t is x2+2x+3
1yes i did try with partial fr as well....
at last we will get a 6 degree eqn in deno..
wat to do with that..??
i am only getting ans with that tanθ wala method... but long method....
62hmm.. i see.. just give me a couple of days.. we are all involved in putting the image uploader :)
so none has too much time to do this as well.. but i promise.. i will be back to my regular quick reply self in a couple of days :)
62actually yeah... right now i cant seem to think of something good..
this is an issue, but i had felt that we could get around it.. but nothing seems to click immediately...
62PHEW.....
Finally got it...
Try integration by parts....
take (x+1)4/(x2+2x+3)3
(x+1)3 . (x+1)/(x2+2x+3)3
Here the 2nd part is (x+1)/(x2+2x+3)3
u will land up with a reduced power .. which u will have to redo using integration by parts...
If this hint doesnt work, i will post the solution :)