[int_(0)^(2x^(2)+3x+3)dx=],[(x+1)(x^(2)+2x+2)^(2)+2x-2ln2-tan],[(pi)/(4)+2ln2-tan^(-1)2quad " b) "(pi)/(4)+2ln x]

int_(0)^(pi//2) log (tan x ) dx=

int_(0)^((pi)/(2))log(tan x)*dx

[int_(0)^(0)cot^(-1)(1-x+x^(2))dx=],[[" (A) "(pi)/(2)-ln2," (B) "(pi)/(2)+ln2],[" (C) "(pi)/(2)-ln sqrt(2)," (D) "(pi)/(2)+ln sqrt(2)]]

int_(0)^( pi/2)log[tan x*cot x]dx

int_(a=)^(oo)ln(x+(1)/(x))(dx)/(1+x^(2))dx=(pi)/(2)ln a then

int_( then )^( If )(1)/(1-tan^(2)x)dx=f(x)+(1)/(4)ln tan((pi)/(4)+x)+c

int_(0)^(pi//2) sin 2 x.ln(tan x)dx

int_(0)^(pi//2) ln (tan x+ cot x)dx=

int_(0)^(3)(3x+1)/(x^(2)+9)dx=(pi)/(12)+log(2sqrt(2))(b)(pi)/(2)+log(2sqrt(2))quad (c)(pi)/(6)+log(2sqrt(2))(d)(pi)/(3)+log(2sqrt(2))

Using integral int_(0)^(-(pi)/(2))ln(sin x)dx=-int_(0)^( pi)ln(sec x)dx=-(pi)/(2)ln2 and int_(0)^((pi)/(2))ln(tan x)dx=0 and int_(0)^((pi)/(4))ln(1+tan x)dx=(pi)/(8)