Integral difficult problem?Any math expert

If integral of g(x) . dx = f(x),then integral of x^3 . g(x^2) is equal to?
(A) (1/2)[x^2* (f(x))^2] - integral of (f(x))^2 . dx
(B) (1/2)[x^2* f(x^2) - integral of f(x^2)* d(x^2)]
(C) (1/2)[x^2* f(x) - (1/2)*integral of (f(x))^2 . dx]
(D) none of these

Note that the answer choices suggest that integration by parts is involved.

If we let u = x^2 ==> du = 2x dx, then the integral becomes:
∫ x^3*g(x^2) dx = 1/2 ∫ x^2*g(x^2) (2x dx)
= 1/2 ∫ u*g(u) du, by applying substitutions.

Integrating this by parts with:
u₀ = u and dv = g(u) ==> du₀ = du and v = ∫ g(u) du = f(u),

gives:
1/2 ∫ u*g(u) = (1/2)(u₀v - ∫ v du₀)
= (1/2)[u*f(u) - ∫ f(u) du].

Therefore, back-substituting u = x^2 gives:
∫ x^3*g(x^2) dx = (1/2)[x^2*f(x^2) - ∫ f(x^2) dx],

which is (B) if you meant dx instead of d(x^2).

I hope this helps!

hi.
use product rule of integration.
before using it apply LIATE
rule.
L-logarithmic
I-inverse
A-algebric
T-trigonometric
E-exponential

best of luck.
since i dont have integral sign in my mobile, i cant explain you full answer.
best of luck.