integrate

∫(x+1)4/(x2+2x+3)3 dx

14 Answers

62
Lokesh Verma ·

does substituting x2+2x+3 work?

just try out

62
Lokesh Verma ·

(x+1) = √t-2

1
skygirl ·

u c...
x2+2x+3=(x+1)2+2
r8?

x+1=t

∫t4/(t2+2)3 dt

then wat???

62
Lokesh Verma ·

after this, the step is t2+2 = k (is it clear why?)

62
Lokesh Verma ·

may be t= √2 tanθ ?

try both i think they will work :)

1
skygirl ·

no 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????

62
Lokesh Verma ·

how 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

1
skygirl ·

yes 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....

62
Lokesh Verma ·

hmm.. 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 :)

62
Lokesh Verma ·

actually 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...

1
skygirl ·

ok fine :)

1
skygirl ·

did u find sumthing simpler????

62
Lokesh Verma ·

PHEW.....

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 :)

1
skygirl ·

ok solved.....
THANK YOU :)....

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