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

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

(i) int((x^(2) - 1)/(x^(2) + 1))dx , (ii) int ((x^(6)- 1)/(x^(2) + 1))dx (iii) int ((x^(4))/(1+x^(2)))dx , (iv) int((x^(2))/(1+x^(2)))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(1)/((1+x^(2))tan ^(-1) x )dx " "(ii) int(e^(tan^(-1)x))/(1+x^(2))dx

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

Evaluate: (i) int((x^(3) + 4x^(2) -3x -2))/((x+2)) dx , (ii) int((x^(4) +1)/(x^(2) +1))dx

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

Evaluate: (i) int(1)/(sin^(3)x cos^(5)x)dx (ii) int(1)/(sin^(3)x cos x)dx

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