`int 1/(3x^2-2x-1) dx`

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

"int(x+2)/(3x^(2)-2x-1)dx

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

(i) int_0^1 (dx)/sqrt(1-x^2) (ii) int_0^1 (dx)/sqrt(1+x^2) (iii) a int_1^sqrt3 (dx)/(1+x^2) b int_0^1 (dx)/(1+x^2) (iv) int_0^(2//3) (dx)/(4+9x^2) (v) int_0^1 x/(x^2+1)dx (vi) int_2^3 x/(x^2+1) dx

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

int(5x-1)/(3x^2+x+2)dx=

int(x^3(2x-1))/(x-1)dx

find int(1)/(x^(2)-a^(2))dx and hence evaluate int(1)/(3x^(2)+13x-10)dx

int(1)/(2x^(2)+3x-4)dx