int frac{f(x) varphi^{prime}(x)+varphi(x) f^{ | vedclass

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(C) Let $u = f(x) \varphi(x)$. Then, by the product rule, $du = (f(x) \varphi^{\prime}(x) + \varphi(x) f^{\prime}(x)) dx$. Substituting this into the

Vedclass · Accepted Answer

$\sqrt{2} 	an ^{-1} \sqrt{\frac{f(x) \varphi(x)-1}{2}}+c$

Vedclass · Answer

(C) Let $u = f(x) \varphi(x)$. Then, by the product rule, $du = (f(x) \varphi^{\prime}(x) + \varphi(x) f^{\prime}(x)) dx$. Substituting this into the integral, we get: $I = \int \frac{du}{(u+1) \sqrt{u-1}}$. Now, let $u-1 = p^2$, which implies $u = p^2 + 1$ and $du = 2p dp$. Substituting these into the integral: $I = \int \frac{2p dp}{(p^2 + 1 + 1) \sqrt{p^2}} = \int \frac{2p dp}{(p^2 + 2) p} = \int \frac{2 dp}{p^2 + 2}$. Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$: $I = 2 \cdot \frac{1}{\sqrt{2}} 	an^{-1}(\frac{p}{\sqrt{2}}) + c = \sqrt{2} 	an^{-1}(\frac{p}{\sqrt{2}}) + c$. Substituting back $p = \sqrt{u-1} = \sqrt{f(x) \varphi(x) - 1}$: $I = \sqrt{2} 	an^{-1} \sqrt{\frac{f(x) \varphi(x) - 1}{2}} + c$.

$\int \frac{f(x) \varphi^{\prime}(x)+\varphi(x) f^{\prime}(x)}{(f(x) \varphi(x)+1) \sqrt{f(x) \varphi(x)-1}} dx=$

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