If `intf(x)dx=psi(x)`, then `intx^5f(x^3")"dx`is equal to(1) `1/3x^3psi(x^3)-3intx^3psi(x^3)dx+C`(2) `1/3x^3psi(x^3)-intx^2psi(x^3)dx+C`(3) `1/3x^3psi(x^3)-intx^3psi(x^3)dx+C`(4) `1/3[x^3psi(x^3)-intx^2psi(x^3)dx]+C`

If `intf(x)dx=psi(x)`, then `intx^5f(x^3")"dx`is equal to(1) `1/3x^3ps

1.	If `intf(x)dx=psi(x)`, then `intx^5f(x^3")"dx`is equal to(1) `1/3x^3psi(x^3)-3intx^3psi(x^3)dx+C`(2) `1/3x^3psi(x^3)-intx^2psi(x^3)dx+C`(3) `1/3x^3psi(x^3)-intx^3psi(x^3)dx+C`(4) `1/3[x^3psi(x^3)-intx^2psi(x^3)dx]+C`
Answer» `I= int x^5(f(x^3)) dx` `` let `x^3 = t` `3x^2 dx = dt` `x^2dx = 1/3 dt` now, `int (t f(t)) dt` `= t int f(t) dt - int (1 int (f(t)) dt) dt` `t psi(**) - int( psi (t)) dt` `x^3 psi(x^3) - 3int x^2 psi (x^3 ) dx` option 2 is correct