Integration by Parts: Step-by-Step Guide
Integration by parts is the integral version of the product rule. It says integral of u dv equals u*v minus integral of v du, and it converts a difficult integral into a (hopefully) easier one. Choose u so that du is simpler than u, and choose dv so that v is easy to find.
Math Racer skill guide · Calculus 2 · AP Calc BC / College
How to solve it, step by step
- Choose u and dv using LIATE. Use the LIATE heuristic to pick u in this priority order: Logarithmic, Inverse trig, Algebraic, Trig, Exponential. Whatever is left, including dx, becomes dv.\[u = \text{(LIATE choice)},\quad dv = \text{(remaining factor)}\,dx\]
- Compute du and v. Differentiate u to get du, and integrate dv to get v (no +C needed at this stage).\[du = u'(x)\,dx,\quad v = \int dv\]
- Apply the parts formula. Substitute into integral of u dv equals u*v minus integral of v du.\[\int u\,dv = uv - \int v\,du\]
- Evaluate the remaining integral. Integrate v du. If it is still hard, you may need parts again or a substitution.\[\int v\,du = V(x) + C\]
- Combine and add +C. Write the final answer u*v - V(x) + C and verify by differentiation.\[\int u\,dv = uv - V(x) + C\]
Worked example
Evaluate the integral of x*e^x dx.
Step 1: By LIATE, the algebraic factor x is u and dv = e^x dx.
Step 2: Compute du = dx and v = integral of e^x dx = e^x.
Step 3: Apply the formula: integral of x e^x dx = x e^x - integral of e^x dx.
Step 4: Evaluate the remaining integral: integral of e^x dx = e^x.
Step 5: Combine: x e^x - e^x + C, which factors as (x - 1) e^x + C.
Verification: Differentiate (x - 1) e^x using the product rule: e^x + (x - 1) e^x = x e^x. Matches.
Tips and common mistakes
- Use LIATE to pick u: Logarithmic, Inverse trig, Algebraic, Trig, Exponential.
- Choose u so that du is simpler; choose dv so that v is easy to find.
- Don't add +C when computing v partway through; only add it at the very end.
- Sometimes you must apply integration by parts twice (e.g., for x^2 e^x).
- For integrals like integral of e^x sin(x) dx, applying parts twice produces the original integral; solve algebraically.
- For integral of ln(x) dx, take u = ln(x) and dv = dx; this is the textbook trick.
- Always verify by differentiating your final answer with the product rule.
- If the new integral looks harder than the original, swap your choice of u and dv.