`intdx/((x-1)^(3/4)(x+2)^(5/4))=` (A) `4/3((x-1)/(x+2))^(1/4)+c` (B) `4/3sqrt((x-1)/(x+2))+c` (C) `4/3((x+2)/(x-1))^(1/4)+c` (D) none of these

intdx/(sqrt(x)(1+x^2)^(5/4))= (A) (2x)/(1+x^2)^(1/4)+c (B) (2sqrt(x))/(1+x^2)^(1/4)+c (C) (2sqrt(x))/(1+x^2)+c (D) none of these

intdx/root(3)((x+1)^2(x-1)^4)

intdx/root(3)((x+1)^2(x-1)^4)

int(x^4-1)/(x^2sqrt(x^4+x^2+1))dx= (a) sqrt(x^2+1/(x^2)+1)+C (b) (sqrt(x^4+x^2+1))/(x^2)+C (c) (sqrt(x^4+x^2+1))/x+C (d) none of these

int ((tan^(-1) x )^(3))/( 1+x^(2)) dx is equal to a) 3 ( tan^(-1) x )^(2) +C b) (1)/(4) ( tan^(-1)x)^(4) +C c) ( tan^(-1) x)^(4) + C d) none of these

d/(dx)[log{e^x((x-2)/(x+2))^(3//4)}] equals (a) (x^2-1)/(x^2-4) (b) 1 (c) (x^2+1)/(x^2-4) (d) e^x(x^2-1)/(x^2-4)

(3x^(2)+x+1)/((x-1)^(4))=(A)/(x-1)+(B)/((x-1)^(2))+(C)/((x-1)^(3))+(D)/((x-1)^(4)) then A+B-C+D=

Evaluate: int(dx)/(x^2(x^4+1)^(3/4)) (A) - ( 1 + 1/x^4)^(1/4) + c (B) ( 1 + 1/x^4)^(1/4) + c (C) - ( 1 - 1/x^4)^(1/4) + c (D) - ( 1 +1/x^4)^(1/4) + c

Evaluate: int(dx)/(x^2(x^4+1)^(3/4)) (A) - ( 1 + 1/x^4)^(1/4) + c (B) ( 1 + 1/x^4)^(1/4) + c (C) - ( 1 - 1/x^4)^(1/4) + c (D) - ( 1 +1/x^4)^(1/4) + c

Integral of the form int[xf\'(x)+f(x)]dx can be evaluated by using integration by parts in first integral and leaving second integral as it is.Now answer the question: int[1/(x^4+1)-(4x^4)/(x^4+1)^2]dx= (A) x^2/(x^4+1)+c (B) x^3/(x^4+1)+c (C) x/(x^4+1)+c (D) none of these