Integration by Parts (Section 7.1): Techniques and Examples

Integration Techniques

Integration by Parts

Integration by parts is a fundamental technique in calculus used to integrate products of functions. It is derived from the product rule for differentiation and is especially useful when the integrand is the product of two functions, one of which becomes simpler when differentiated and the other when integrated.

• Definition: If u and v are differentiable functions of x, then the integration by parts formula is:

$\int u \, dv = uv - \int v \, du$

• Purpose: To transform the integral of a product of functions into (hopefully) simpler integrals.

• Strategy: Choose u to be a function that becomes simpler when differentiated, and dv to be a function that is easy to integrate.

Conditions for Using Integration by Parts

Integration by parts can be applied only if the following conditions are satisfied:

• The integrand can be written as the product of two functions, u and dv.

• It is possible to integrate dv to get v and to differentiate u to get du.

• The integral $\int v \, du$ can be found.

Examples of Integration by Parts

• Example 1: Find $\int x e^x \, dx$.

  • Let $u = x$, $dv = e^x dx$.

  • Then $du = dx$, $v = e^x$.

  • Apply the formula:

  • $\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C$

• Example 2: Find $\int \ln x \, dx$ for $x > 0$.

  • Let $u = \ln x$, $dv = dx$.

  • Then $du = \frac{1}{x} dx$, $v = x$.

  • Apply the formula:

  • $\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - \int 1 dx = x \ln x - x + C$

• Example 3: Find $\int (2x + 5) e^{-x} dx$.

  • Let $u = 2x + 5$, $dv = e^{-x} dx$.

  • Then $du = 2 dx$, $v = -e^{-x}$.

  • Apply the formula:

  • $\int (2x + 5) e^{-x} dx = (2x + 5)(-e^{-x}) - \int -e^{-x} \cdot 2 dx$

  • Simplify and integrate further as needed.

• Example 4: Find $\int x^2 \ln x \, dx$.

  • Let $u = \ln x$, $dv = x^2 dx$.

  • Then $du = \frac{1}{x} dx$, $v = \frac{x^3}{3}$.

  • Apply the formula:

  • $\int x^2 \ln x dx = \ln x \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} dx$

  • Simplify $\int \frac{x^3}{3} \cdot \frac{1}{x} dx = \int \frac{x^2}{3} dx = \frac{x^3}{9}$

  • Final answer: $\int x^2 \ln x dx = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C$

Guidelines for Choosing u and dv

• Use the LIATE rule to help choose u (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential).

• Choose u as the function that becomes simpler when differentiated.

• Choose dv as the function that is easy to integrate.

Summary Table: Integration by Parts

Step	Description
1	Identify u and dv in the integrand.
2	Compute du and v.
3	Apply the formula $\int u \, dv = uv - \int v \, du$.
4	Simplify and solve the resulting integral.

Additional info: The examples provided illustrate both single and repeated use of integration by parts, as well as its application to logarithmic, exponential, and polynomial functions.