If int f(x) dx = psi(x),then int x^5 f(x^3) d | vedclass

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(A) Given $\int f(x) dx = \psi(x)$.Let $I = \int x^5 f(x^3) dx = \int x^3 f(x^3) x^2 dx$.Substitute $t = x^3$,then $dt = 3x^2 dx$,which implies $x^2 d

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$\frac{1}{3}[x^3 \psi(x^3)] - \int x^2 \psi(x^3) dx$

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(A) Given $\int f(x) dx = \psi(x)$.Let $I = \int x^5 f(x^3) dx = \int x^3 f(x^3) x^2 dx$.Substitute $t = x^3$,then $dt = 3x^2 dx$,which implies $x^2 dx = \frac{dt}{3}$.Substituting these into the integral,we get $I = \int t f(t) \frac{dt}{3} = \frac{1}{3} \int t f(t) dt$.Using integration by parts,where $u = t$ and $dv = f(t) dt$ (so $du = dt$ and $v = \psi(t)$):$I = \frac{1}{3} [t \psi(t) - \int \psi(t) dt]$.Now,substitute $t = x^3$ back into the expression:$I = \frac{1}{3} [x^3 \psi(x^3) - \int \psi(x^3) (3x^2 dx)]$.$I = \frac{1}{3} x^3 \psi(x^3) - \int x^2 \psi(x^3) dx$.

If $\int f(x) dx = \psi(x)$,then $\int x^5 f(x^3) dx = $

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