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Vol. I — Calculus
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This is a real StudyPrime answer to a first-year calculus integral — steps labeled with the why, the check included, the final answer boxed. The point isn't the answer. It's that you could do the next one yourself.

QuestionEvaluate ∫xex dx\displaystyle\int x e^x \, dx
Step 1 — Choose u and dv

Integration by parts. Pick the part that gets simpler when differentiated: u=xu = x and dv=ex dxdv = e^x\,dx, so du=dxdu = dx and v=exv = e^x.

Step 2 — Apply the formula
∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du
∫xex dx=xex−∫ex dx\int x e^x\,dx = x e^x - \int e^x\,dx
Step 3 — Integrate what's left
=xex−ex+C=(x−1)ex+C= x e^x - e^x + C = (x - 1)e^x + C
Check — differentiate the answer

ddx[(x−1)ex]=ex+(x−1)ex=xex\tfrac{d}{dx}\big[(x-1)e^x\big] = e^x + (x-1)e^x = x e^x ✓ matches the integrand

Final answer (x−1)ex+C \boxed{\,(x-1)e^x + C\,}
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