IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

aakash.pk

integrate 1/(e^x+1)

Aatish

i am trying to do it.let's communicate

rajatkashyap

_{[ ]}

^{[ ]} 1/(e^x+1)dx

TAKE e^x COMMON FROM DENOMINATOR,

_{[ ]}

^{[ ]} e^(-x)/(1+e^(-x))dx

TAKE ,

1+e^(-x)=t

differentiating both sides w.r.t. x;

-e^(-x)dx=dt

putting in eqn.

_{[ ]}

^{[ ]} -dt/t

-ln(t)+c

ln(1/t)+c

putting the value of t

ln(1/(1+e^(-x))+c

ln(e^x/(1+e^x))+c

i hope this is correct!!

Aatish

will u give me a little hint I am very near to your answer

pritambhalerao

x-ln[e^x+1]+c,

this is acoording to,

numerator=a[denominator]+b[derivativeof denominator]

by this method it quit esay.

nachiket

1/e^x+1=e^x/e^x(1+e^x)=

dt/(t-1)t

using partial fractions:

1=At+B(t-1)

so,A=1&B=-1

so,

dt/(t-1)-

dt/t=ln(t-1)-lnt+c=ln{(t-1)/t}+c=ln{e^x/(e^x+1)}+c