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

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[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[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

Method of integration by parts : int(x-sinx )/(1-cosx)dx=....

Integrate : int x tanx sec^(2)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

Find the integral int(a x^2+b x+c)dx

Integrate: int (cos2x)/(cosx+sinx)dx

Method of integration by parts : inte^(x)[tanx-log(cosx)]dx=.....