IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

v_gurucharan

(LOG (X²+1)- LOG(X))(1/1+X²) dx

avi_1214545

edited..

elastiboysai

i tried using parts.

log(x^2+1/x)(1/x^2+1)dx
log(x^2+1/x)*tan^-1x-

tan^-1x*x^2-1/(x^2+1 *x) dx
now make the substitution x =tan y
simplify n uget

y[2coty-cosecysecy]dy
y coty and ysecycosecy r not simple integrals.
but they r evaluable.

..ill try for oder methods..

computer001

integral{ log((x^2+1)/x)*1/1+x^2 dx}
x=tany
integral{ log{sec^2ycoty}dy
= -1*integral{log(sinycosy)}dy
so u can simplify tht 2 integral log{sint}dt + some direct stuff
log(sint)dt is direct with definite...r u sure its indefinite??

v_gurucharan

yes its indefinite!!!!@@!!!!!

@ elastiboysai (2007)
ur sol sems to be correct!!
but i don't think a question from a cbse buk will be so lenghty for 4mm

elastiboysai

gurucharan,
are u sure u got this frm the cbse text book?
this doesn seem simple,
konichiwa also tried it n said he 2 got somthin similiar to wat i have got.
cn u chek up ur qn plz?

computer001

elasti, mine??
n how do u integrate log(sinx) indef...
i dunno its not comin easily

elastiboysai

ur methods also rt..
and

logsinxdx cant be evaluated easily using parts
i hav tried it b4 but in vain.

elastiboysai

Now that u hv changed it..
_{[0 ]}

^{[infinity ]} log(x^2+1)/x *1/x^2+1 dx
now substitute x=tan a
_{[0 ]}

^{[pi/2 ]}log(sec^2a/tana)da
_{[0 ]}

^{[pi/2 ]}log(1/sina)+log(1/cosa) da
shud be pi *log2
using the fact that
_{[0 ]}

^{[pi/2 ]}logsina da= -pi/2 *log2

v_gurucharan

but i hav not changed the question!!!
this is question with limits is given on same page of buk!!

elastiboysai

hey i meant u changed the qn from definite to indefinite
the integral is same anyways

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