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

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(C) Given $\int f(x) dx = \psi(x)$.Let $I = \int x^5 f(x^3) dx$.We can rewrite this as $I = \int x^3 \cdot x^2 f(x^3) dx$.Let $x^3 = t$,then $3x^2 dx

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

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(C) Given $\int f(x) dx = \psi(x)$.Let $I = \int x^5 f(x^3) dx$.We can rewrite this as $I = \int x^3 \cdot x^2 f(x^3) dx$.Let $x^3 = t$,then $3x^2 dx = dt$,which implies $x^2 dx = \frac{1}{3} dt$.Substituting these into the integral:$I = \int t \cdot f(t) \cdot \frac{1}{3} dt = \frac{1}{3} \int t f(t) dt$.Using integration by parts,$\int u dv = uv - \int v du$,where $u = t$ and $dv = f(t) dt$:$I = \frac{1}{3} [t \int f(t) dt - \int (\frac{d}{dt}(t) \cdot \int f(t) dt) dt] + c$.Since $\int f(t) dt = \psi(t)$:$I = \frac{1}{3} [t \psi(t) - \int \psi(t) dt] + c$.Substituting $t = x^3$ and $dt = 3x^2 dx$:$I = \frac{1}{3} [x^3 \psi(x^3) - \int \psi(x^3) \cdot 3x^2 dx] + c$.$I = \frac{1}{3} [x^3 \psi(x^3) - 3 \int x^2 \psi(x^3) dx] + c$.$I = \frac{1}{3} x^3 \psi(x^3) - \int x^2 \psi(x^3) dx + c$.

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

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