Integration by Parts

Concept Explanation

Integration by parts converts a difficult product integral into a simpler one. To select $u$ effectively, use the LIATE mnemonic priority (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential). You differentiate $u$ to get $du$, and integrate $dv$ to get $v$.

Visual / Geometric Intuition

Interactive Desmos Graph

Engineering Applications

Extensively applied in signal analysis to solve Fourier series coefficients, evaluating center of mass integrals, and modeling mechanical dampening parameters in vibration analysis.

Example Problem

Evaluate:

Solution

Using LIATE, we select $u = \ln x$ (logarithmic) and $dv = x \, dx$ (algebraic).

Step 1: Compute derivatives and integrals

• $u = \ln x \implies du = \frac{1}{x} \, dx$
• $dv = x \, dx \implies v = \frac{1}{2} x^2$

Step 2: Apply the formula

$$\int x \ln x \, dx = (\ln x)\left(\frac{1}{2} x^2\right) - \int \left(\frac{1}{2} x^2\right)\left(\frac{1}{2x} \, dx\right)$$

$$= \frac{1}{2} x^2 \ln x - \frac{1}{2} \int x \, dx$$

$$= \frac{1}{2} x^2 \ln x - \frac{1}{2}\left(\frac{1}{2} x^2\right) + C$$

$$= \frac{1}{2} x^2 \ln x - \frac{1}{4} x^2 + C$$

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Connections & References

• Parent Concepts: Substitution Rule
• Sub-concepts: Improper Integrals
• Course Links: Calc-2 Session 14