IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

v_gurucharan

INTEGRATE {(1+X²)^1/2}/(1-X²) WRT X !!!

I FIND INTEGRATION DIFFICULT , PLS. TELL ME IS THERE ANY WAY TO COPE WITH THE PROB.???

I HAV ALREADY DONE AROUND 100 PROB. OF INTEGRATION

studyid

hi guru,

never look at how many probs u hav done............

i will never help u........

regarding yr prob...

just wait for five mins..........

i am lookin at it

studyid

hey guru really a nice sum...

still need some minss....
hope u dont mind...

akhil_o

put ^{[ ]}

1+x²=t

in

1+x²

--------

(1-x²)

1+x²

^{=-2 t²dt
        ----------
        2-t²

=-2 4 - t²dt - 4       dt
          ----------       ----------
           2-t^{2          2-t²}

from this we get}

=

2t - 2

2{ln (2+t)/(2-t)} +c

Resubstitute t =(1+x²)^1/2

to get the answer

hope it helps

studyid

this is what i hav done...

see...
take out x²
out from the numerator and x from the denominator

so u have rt(1+1/x²)/x-1/x

now if u watch then the derivative of 1-1/x is set in th numeratoe.......

but the prob. arising with me is that the num. is in root..

how to kill that root....

kaushalbharat

hi , divide numerator&denominator by x
then takethe denominator as t
and dt will be numerator
rate me if this is use full

akhil_o

anything wrong in my answer?

raulrag009

multiply root(1+x²) in numerator and denominator

and then continue.

it becomes

[(1+x²)/(1-x²)

1+x²]dx

now solving , this by separating the numerator each with the denominator

then it will become easy!!!!!!

studyid

hey akhil good way out...

but still i am confused with my method or can there be some trigo subs.....

akhil_o

no i tried that but nuthin was workin out...the one plus and one minus holds us up there i guess

hsbhatt

trig substitution that works t = tan

.

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