If int e^x(1+x) cdot sec ^2(x e^x) , dx = f(x | vedclass

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(D) Given the integral: $\int e^x(1+x) \cdot \sec^2(x e^x) \, dx = f(x) + C$. Let $t = x e^x$. Then, differentiating with respect to $x$ using the pro

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$	an(x e^x)$

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(D) Given the integral: $\int e^x(1+x) \cdot \sec^2(x e^x) \, dx = f(x) + C$. Let $t = x e^x$. Then, differentiating with respect to $x$ using the product rule: $\frac{dt}{dx} = e^x + x e^x = e^x(1+x)$. Thus, $dt = e^x(1+x) \, dx$. Substituting these into the integral, we get: $\int \sec^2(t) \, dt$. The integral of $\sec^2(t)$ is $	an(t) + C$. Substituting back $t = x e^x$, we get $	an(x e^x) + C$. Comparing this with $f(x) + C$, we find $f(x) = 	an(x e^x)$.

If $\int e^x(1+x) \cdot \sec ^2(x e^x) \, dx = f(x) + \text{constant}$, then $f(x)$ is equal to

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