I=``int 1/(3+2x+x^2)dx`

I=int(1)/(x^(3)+x^(2)+x+1)dx

I= int (dx)/( x^(2) + x+1)

(i) int 1/(x^2+2x+2) dx (ii) int 1/(9x^2+6x+5) dx

(i) int 1/(1-x^2) dx (ii) int 1/(x^2-9) dx

If I_(1)=int_(0)^(1) 2^(x^(2)) dx, I_(2)=int_(0)^(1) 2^(x^(3)) dx, I_(3)=int_(1)^(2) 2^(x^(2))dx and I_(4)=int_(1)^(2) 2^(x^(2))dx then

I = int( 6 x^(3) + x^(2)-2x + 1) /( 2x - 1) dx .

(i) int_1^3 (3x)/(9x^2-1) dx (ii) int_2^3 x/(x^2+1) dx

(i) int (3x^2)/(x^6+1) dx (ii) int (3x)/(1+2x^4) dx

(i) int x^3 sin x^2 dx (ii) int x^3 cos x^2 dx