If int f(x) dx = g(x), then int x^3 f(x^2) dx | vedclass

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(A) Let $I = \int x^3 f(x^2) dx$. Substitute $t = x^2$, then $dt = 2x dx$, which implies $x dx = \frac{1}{2} dt$. Also, $x^2 = t$. Substituting these

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$\frac{1}{2} [x^2 g(x^2) - \int g(x^2) d(x^2)]$

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(A) Let $I = \int x^3 f(x^2) dx$. Substitute $t = x^2$, then $dt = 2x dx$, which implies $x dx = \frac{1}{2} dt$. Also, $x^2 = t$. Substituting these into the integral: $I = \int x^2 f(x^2) (x dx) = \int t f(t) \frac{1}{2} dt = \frac{1}{2} \int t f(t) dt$. Using integration by parts, $\int u dv = uv - \int v du$, let $u = t$ and $dv = f(t) dt$. Then $du = dt$ and $v = \int f(t) dt = g(t)$. Thus, $I = \frac{1}{2} [t g(t) - \int g(t) dt]$. Substituting $t = x^2$ back, we get $I = \frac{1}{2} [x^2 g(x^2) - \int g(x^2) d(x^2)]$.

If $\int f(x) dx = g(x)$, then $\int x^3 f(x^2) dx$ is equal to

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