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Integration by Parts (IBP): The Ultimate Guide with Practice Problems

IBP is the reverse of the product rule for integration.

Core Theorem
∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du

Practice Exercises

Example 01Medium
Evaluate ∫xex dx\int x e^x \, dx.
NEED A HINT?
Use LIATE: xx is Algebraic (choose as uu), exe^x is Exponential (choose as dvdv).
SHOW DETAILED EXPLANATION

Step 1: Assign uu and dvdv

u=x,dv=exdxu = x, dv = e^x dx.

Step 2: Find dudu and vv

du=dx,v=exdu = dx, v = e^x.

Step 3: Apply Formula

xex−∫ex dx=xex−ex+Cxe^x - \int e^x \, dx = xe^x - e^x + C.
Example 02Easy
Evaluate ∫ln⁡x dx\int \ln x \, dx.
NEED A HINT?
Treat ln⁡x\ln x as ln⁡x⋅1\ln x \cdot 1. Use LIATE: ln⁡x\ln x is Logarithmic (choose as uu).
SHOW DETAILED EXPLANATION

Step 1: Assign uu and dvdv

u=ln⁡x,dv=dxu = \ln x, dv = dx.

Step 2: Find dudu and vv

du=1xdx,v=xdu = \frac{1}{x} dx, v = x.

Step 3: Apply Formula

xln⁡x−∫x⋅1x dx=xln⁡x−x+Cx \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C.
Example 03Hard
Evaluate ∫x2cos⁡x dx\int x^2 \cos x \, dx.
NEED A HINT?
You will need to apply IBP twice (or use the Tabular Method). Choose u=x2u = x^2 to reduce its degree.
SHOW DETAILED EXPLANATION

Step 1: First IBP

Let u=x2,dv=cos⁡xdx  ⟹  du=2xdx,v=sin⁡xu = x^2, dv = \cos x dx \implies du = 2x dx, v = \sin x. Result: x2sin⁡x−∫2xsin⁡x dxx^2 \sin x - \int 2x \sin x \, dx.

Step 2: Second IBP

For ∫2xsin⁡x dx\int 2x \sin x \, dx, let u=2x,dv=sin⁡xdx  ⟹  du=2dx,v=−cos⁡xu = 2x, dv = \sin x dx \implies du = 2 dx, v = -\cos x.

Step 3: Final Consolidation

x2sin⁡x−(−2xcos⁡x+2sin⁡x)=x2sin⁡x+2xcos⁡x−2sin⁡x+Cx^2 \sin x - (-2x \cos x + 2 \sin x) = x^2 \sin x + 2x \cos x - 2 \sin x + C.
Common Pitfalls
  • ⚠
    Wrong 'u' SelectionChoosing uu based on what looks 'easy' rather than LIATE can lead to an even more complex integral. For ∫xln⁡x dx\int x \ln x \, dx, picking u=xu=x makes it harder; always prioritize Logarithms as uu.
  • ⚠
    Nested Sign ErrorsWhen performing multiple steps of IBP, it's very easy to lose track of the negative signs from the formula ∫v du\int v \, du, especially when vv itself contains a negative sign (like −cos⁡x-\cos x).
  • ⚠
    The +C Missing LinkIBP is an indefinite integration technique. Forgetting to add the constant of integration (+C+C) after the final step is a common point deduction in exams.
Gary Chang

Gary Chang

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5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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