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

Options

A. `f(x^(2))g'(x^(2)) - g(x^(2))f'(x^(2))+c`
B. `(1)/(2) {f(x^(2)) g'(x^(2)) - g(x^(2))f(x^(2))}+c`
C. `(1)/(2) {f(x^(2))g'(x^(2))- g(x^(2))f'(x^(2))}+c`
D. None of the above

Answer

C

Explanation

Let `I = int xx{f(x^(2)) g'' (x^(2)) - f''(x^(2)) g(x^(2))} dx`
Put `x^(2) = t rArr 2x dx = dt`
` therefore I = (1)/(2) [ int {f(t) g''(t)]dt - int f'' (t)g'' (t)dt]`
` = (1)/(2) [f(t)g'(t) - int f'(t)g'(t)dt - g(t') f'(t) + int g' (t) f'(t) dt] + c`
` = (1)/(2)[f(t) g'(t) - g(t)f'(t)]+c`
` = (1)/(2)[f(x^(2)) g'(x^(2)) - g(x^(2)) f'(x^(2))] + c`