IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

chinmay_saxena01

$\int1/{x}-{x}^{3}dx$ .....

the answer is `+c`

computer001

wat is ur doubt?

chinmay_saxena01

i will tell u later ,,,first solve it....tumhare sol. me se hi apna doubt bata doonga,,,

computer001

well

write as dx/x(1-x^2) =dx/x(1-x)(1+x)

[1/3{ 2/x(1+x) + 1/(1-x)(1+x) + 1/x(1-x)}]dx

= {1/x -1/2*1/1+x + 1/2*1/1-x}dx

intergrating v get:

lnx-1/2ln(1+x) -1/2ln(1-x) +c

= 1/2{ 2lnx - ln 1-x^2} +c

=1/2{ln(x^2/1-x^2)} +c

chinmay_saxena01

ok now my method....1/x-x^3=1/x(1-x^2)
so now let logx = t,,,so we will get i/xdx=dt,,,
so dt/(1-e^2t) ,,,now we can integrate this by using .....dx/a^2-x^2=1/2loga+x/a-x+c,,,,,
so the answer i m getting is 1/2log(1+x/1-x)+c...is it correct,,,?

computer001

dt(1-e^2t) is not of the form dt/a^2-t^2

note tht in formula the variable occurs as a normal term...in ur method it apperas as a power...so u cant use the formula..it aint connected to the sum

MUDIT

HERE IS ANOTHER WAY
dx/x-x^3 multiply and divide by x so we have xdx/x^2-x^4 and put x^2=t so dx=dt/2x and hence the expression reduces to
1/2[dt/t-t^2] = 1/2[dt/t(1-t)] = 1/2[dt/t+dt/1-t] = 1/2[lnt-ln(1-t)]
=1/2[lnt/1-t]+c substitute t=x^2 and you get the required ans

chinmay_saxena01

oh oh oh oh,,,silly mistake,,,,neways thanks very much,,,so i have rated u,,,

computer001

cheers!! rates aint important to me...as long as u understood im satisfied :)

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the answer is +c

the answer is `+c`