30" The value of "int[f(x)g''(x)-f''(x)g(x)]dx" is equal to "

int{f(x)g'(x)-f'g(x)}dx equals

Let f(x) be a function satisfying f'(x)=f(x) with f(0)=1 and g(x) be a function that satisfies f(x)+g(x)=x^(2) . Then the value of the integral int_(0)^(1) f(x)g(x) dx is equal to -

If an antiderivative of f(x) is e^(x) and that of g(x) is cosx, then the value of int f(x) cos x dx+ int g(x) e^(x) dx is equal to -

If (d)/(dx)f(x)=g(x) , then the value of int_(a)^(b)f(x)g(x)dx is -

The value of int_1^2 {f(g(x))}^(-1)f'(g(x))g'(x) dx , where g(1)=g(2), is equal to

The value of int_1^2 {f(g(x))}^(-1)f'(g(x))g'(x) dx , where g(1)=g(2), is equal to

int x{f(x)^(2) g''(x^(2))-f''(x^(2)) g(x^(2))} dx is equal to

If g(1)=g(2), then int_(1)^(2)[f{g(x)}]^(-1)f'{g(x)}g'(x)dx is equal to