`(xe^x)/(1+x)^2dx=` (A) `e^x/(1+x)` (B) `e^x/(1+x)^2` (C) `e^xlog(1+x)` (D) none of these

The integral inte^x(f(x)+f\'(x))dx can be solved by using integration by parts such that: I=inte^xf(x)dx+inte^xf\'(x)dx=e^xf(x)-inte^xf\'(x)dx+inte^xf\'(x)dx=e^xf(x)+C , and inte^(ax)(f(x)+(f\'(x))/a)dx=e^(ax)f(x)/a+C ,Now answer the question: int(e^x(2-x^2))/((1-x)sqrt(1-x^2))dx (A) e^xsqrt((1-x)/(1+x))+C (B) e^xsqrt((1+x)/(1-x))+C (C) e^xsqrt((2-x)/(2+x))+C (D) none of these

The differential coefficient of f((log)_e x) with respect to x , where f(x)=(log)_e x , is (a) x/((log)_e x) (b) 1/x(log)_e x (c) 1/(x(log)_e x) (d) none of these

int e^x((x^2+1))/((x+1)^2)dx is equal to ((x-1)/(x+1))e^x+c (b) e^x((x+1)/(x-1))+c e^x(x+1)(x-1)+c (d) none of these

The value of int_1^e((tan^(-1)x)/x+(logx)/(1+x^2))dx ,is (a) tane (b) tan^(-1)e (c) tan^(-1)(1/e) (d) none of these

int e^x ((1-sinx)/(1-cosx)) dx= (A) e^xtan(x/2)+C (B) e^xcot(x/2)+C (C) -1/2 e^xtan(x/2)+C (D) -1/2 e^xcot(x/2)+C

The maximum value of x^(1/x),x >0 is (a) e^(1/e) (b) (1/e)^e (c) 1 (d) none of these

The maximum value of x^(1/x),x >0 is (a) e^(1/e) (b) (1/e)^e (c) 1 (d) none of these

If int_0^1 xe^(x^2) dx=k int_0^1 e^(x^2) dx , then (A) kgt1 (B) 0ltklt1 (C) k=1 (D) none of these

Integration of 1/(1+((log)_e x)^2) with respect to (log)_e x is (tan^(-1)((log)_e x)/x)+C (b) tan^(-1)((log)_e x)+C (c) (tan^(-1)x)/x+C (d) none of these

If f(x)=(sin^(-1)x)/(sqrt(1-x^2)),t h e n(1-x^2)f^(x)-xy(x)= 1 (b) -1 (c) 0 (d) none of these