If int phi(x) dx = psi(x), then int (phi circ | vedclass

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(B) Given $\int \phi(x) dx = \psi(x)$. We need to evaluate $I = \int \phi(h(x)) \cdot h(x) h'(x) dx$. Let $h(x) = t$, then $h'(x) dx = dt$. The integr

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$(\psi \circ h)(x) h(x) - \int (\psi \circ h)(x) h'(x) dx + c$

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(B) Given $\int \phi(x) dx = \psi(x)$. We need to evaluate $I = \int \phi(h(x)) \cdot h(x) h'(x) dx$. Let $h(x) = t$, then $h'(x) dx = dt$. The integral becomes $I = \int \phi(t) \cdot t dt$. Using integration by parts, let $u = t$ and $dv = \phi(t) dt$. Then $du = dt$ and $v = \int \phi(t) dt = \psi(t)$. Applying the formula $\int u dv = uv - \int v du$: $I = t \psi(t) - \int \psi(t) dt$. Substituting $t = h(x)$ back into the expression: $I = h(x) \psi(h(x)) - \int \psi(h(x)) h'(x) dx + c$. This can be written as $(\psi \circ h)(x) h(x) - \int (\psi \circ h)(x) h'(x) dx + c$.

If $\int \phi(x) dx = \psi(x)$, then $\int (\phi \circ h)(x) \cdot h(x) h'(x) dx =$

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