[25.(i)int(2)/((1-x)(1+x^(2)))dx],[(N.C.E.R.T.;C.B.S.E.2]

I=int(e^(2x)-1)/(e^(2x))dx

(i) int (dx)/(e^x+e^(-x)) (ii) int e^x/(e^(2x)+1) dx

If I=int_(0)^(1) (1+e^(-x^2)) dx then, s

If I=int_(0)^(1) (1+e^(-x^2)) dx then, s

int 5^(x+1)*e^(2x-1)dx=....+c

[int_(0)^( pi/2)cos x*e^(sin x)dx=.........],[[" (A) "e+1," (B) "e-1," (C) "e," (D) "-e]]

Show that (a) int_(e)^(e^(2))(1)/(log x)dx = int_(1)^(2)(e^(x))/(x)dx (b) int_(t)^(1)(dx)/(1+x^(2)) = int_(1)^(1//t)(dx)/(1+x^(2))

Statement -1 : If I_(1)=int(e^(x))/(e^(4x)+e^(2x)+1)dx and I_(2)=int(e^(-x))/(e^(-4x)+e^(-2x)+1)dx , then I_(2)-I_(1)=(1)/(2)log((e^(2x)-e^(x)+1)/(e^(2x)+e^(x)+1))+C where C is an arbitrary constant. Statement -2 : A primitive of f(x) =(x^(2)-1)/(x^(4)+x^(2)+1) is (1)/(2)log((x^(2)-x+1)/(x^(2)+x+1)) .

Statement -1 : If I_(1)=int(e^(x))/(e^(4x)+e^(2x)+1)dx and I_(2)=int(e^(-x))/(e^(-4x)+e^(-2x)+1)dx , then I_(2)-I_(1)=(1)/(2)log((e^(2x)-e^(x)+1)/(e^(2x)+e^(x)+1))+C where C is an arbitrary constant. Statement -2 : A primitive of f(x) =(x^(2)-1)/(x^(4)+x^(2)+1) is (1)/(2)log((x^(2)-x+1)/(x^(2)+x+1)) .

Find (i) inte^x(t a n^(-1)x+1/(1+x^2))dx (ii) int((x^2+1)e^x)/((1+x)^2))dx