`int(x^6+1)/(x^2+1)dxi se q u a l` `(x^5)/5+(x^3)/3+x+c` (b) `(x^5)/5-(x^3)/3=x+c` `(x^5)/5-(x^3)/3+(x^2)/2+c` (d) `(x^5)/5-(x^3)/3+x+c`

(2x)/(5-3x)=(5+3x)/(8x)

int(x^(6)-1)/(x^(2)+1)dx=(x^(5))/(5)-(x^(3))/(3)+x+k tan^(-1)x+C then k=

int(x+(1)/(x))^(3/2)((x^(2)-1)/(x^(2)))dx is equal to (A) (1)/(3)(x+(1)/(x))^(3)+C (B)(2)/(5)(x+(1)/(x))^(5/2)

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

(x^(2)+5x+7)/((x-3)^(3))=(A)/(x-3)+(B)/((x-3)^(2))+(C)/((x-3)^(3))rArr A=

(x-1)/(3)+(2x+5)/(6)=(3x-6)/(9)-(2x-5)/(2)

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

int(x^((3)/(2))+(1)/(sqrt(x)))dx is equal to (2)/(5)x^((5)/(2))+2sqrt(x)+c (b) (5)/(2)x^((5)/(2))+2sqrt(x)+c(2)/(5)x^((5)/(2))-2sqrt(x)+c(d)(1)/(2)x^((3)/(2))+2sqrt(x)+c

((1)/(x-3)-(3)/(x(x^(2)-5x+6)))

int (2x ^ (3) + 3x ^ (2) + 4x + 5) / (2x + 1) dx equls to: (A) (x ^ (3)) / (2) + (x ^ (2)) / (2) +3 (x) / (2) + (7) / (4) ln (2x + 1) + C (B) 5 (x ^ (3)) / (2) + 6 (x) / (2) + (7) / (4) ln (2x + 1) + C (C) 3 (x ^ (3)) / (2) + 3 (x ^ (2)) / (2) + (7 ) / (4) ln (2x + 1) + C (D) (x ^ (2)) / (2) + 3 (x) / (2) + (7) / (4) ln (2x + 1) + C