Question

integration 1/1+tangentx

Answer 1

step 1:- let tanx=z
now differentiate
sec^2x .dx=dz
(1+tan^2x)dx=dz
(1+z^2) dx=dz
put dx=dz/(1+z^2)

step 2:-
integ dz/(1+z)(1+z^2)

step 3:-use concept of partial fraction.
i.e dz/(1+z)(1+z^2)=A/(1+z)+(Bz+C)

1=A (1+z^2)+(Bz+C)(1+z)
=A+Az^2+Bz+Bz^2+C+Cz
=(A+B)z^2+(B+C) z+(A+C)
now compare LHS to RHS
A+B=0
B+C=0
A+C=1
A =1/(1+(-1)^2)=1/2
C=-1/2
B=1/2
put its above

step 4:-now integrate
1/2.dz/(1+z) +(1/2z-1/2)dz/(1+z^2)
=1/2ln (1+z)+integ1/2 (z-1)/(1+z^2)
=1/2ln (1+z)+1/4ln (1+z^2)-1/2integ 1/(1+z^2)
=1/2ln (1+z)+1/4ln (1+z^2)-1/2tan^-1 z

now put z=tan^-1(x)