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[x^(x+1)(logx+1)+x^x]dx=` (A) `x^(x+2)+c` (B) `x^(x+1)+c` (C) `x^x+x+c` (D) none of these

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

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(x+sinx)/(1+cosx)dx= (A) xtanx/2+c (B) x/2tanx/2+c (C) x/2tanx+c (D) none of these

Integrate : int(x)/(a +x)dx

If f(x) is an integrable function and f^-1(x) exists, then int f^-1(x)dx can be easily evaluated by using integration by parts. Sometimes it is convenient to evaluate int f^-1(x)dx by putting z=f^-1(x) .Now answer the question.If I_1=int_a^b[f^2(x)-f^2(a)]dx and I_2=int_(f(a))^(f(b)) 2x[b-f^-1(x)]dx!=0 , then I_1/I_2= (A) 1:2 (B) 2:1 (C) 1:1 (D) none of these

Integrate : int(x)/(a^(2)+x^(2))dx

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

Integrate : int ( x / sqrt( 1 + x^2 ) ) dx

Integrate : int(1+x)/(1-2x-x^(2))dx

Integrate : int(sec^(2)x)/(sqrt(1+tan x))dx

Integrate int(x)/(1+x tan x)dx