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If $\int \phi(x) d x=\psi(x)$, then $\int\left(\phi_0 h\right)(x) h(x) h^{\prime}(x) d x=$
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The correct answer is:
$\left(\psi_0 h\right)(x) h(x)-\int\left(\psi_0 h\right)(x) h^{\prime}(x) d x+c$
$\int \phi(x) d x=\psi(x)$, then $\int(\phi \circ h)(x) \cdot h(x) h^{\prime}(x) d x=$ ?
Now, $\int \phi\{h(x)\} \cdot h(x) \cdot h^{\prime}(x) d x$
Let $h(x)=t \Rightarrow h^{\prime}(x) d x=d t$
Now, $\int_{\mathrm{II}}^{\phi(t) \cdot{ }_{\mathrm{I}}} d t$
$=t \cdot \int \phi(t) d t-\int\left\{\frac{d}{d t}(t) \cdot \int \phi(t) d t\right] d t=t \cdot \psi(t)-\int \psi(t) d t$
$=h(x) \psi_0 h(x)-\int \psi_0 h(x) \cdot h^{\prime}(x) \cdot d x$
Now, $\int \phi\{h(x)\} \cdot h(x) \cdot h^{\prime}(x) d x$
Let $h(x)=t \Rightarrow h^{\prime}(x) d x=d t$
Now, $\int_{\mathrm{II}}^{\phi(t) \cdot{ }_{\mathrm{I}}} d t$
$=t \cdot \int \phi(t) d t-\int\left\{\frac{d}{d t}(t) \cdot \int \phi(t) d t\right] d t=t \cdot \psi(t)-\int \psi(t) d t$
$=h(x) \psi_0 h(x)-\int \psi_0 h(x) \cdot h^{\prime}(x) \cdot d x$
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