1. The problem is to find the integral of $x\sec^2(x)\,dx$. 2. We use integration by parts formula: $$\int u\,dv = uv - \int v\,du$$. 3. Let $u = x$ so that $du = dx$. 4. Let $dv = \sec^2(x)\,dx$ so that $v = \tan(x)$ because $\frac{d}{dx}(\tan x) = \sec^2 x$. 5. Applying integration by parts: $$\int x\sec^2(x)\,dx = x\tan(x) - \int \tan(x)\,dx$$ 6. The integral of $\tan(x)$ is: $$\int \tan(x)\,dx = -\ln|\cos(x)| + C$$ 7. Substitute back: $$\int x\sec^2(x)\,dx = x\tan(x) + \ln|\cos(x)| + C$$ 8. This is the final answer.