Integration by Parts (Section 7.1): Techniques and Examples

7.1 Integration Techniques

Integration by Parts

Integration by parts is a fundamental technique for evaluating integrals where the integrand is a product of two functions. It is especially useful when standard integration methods do not apply directly.

• Definition: Integration by parts is based on the product rule for differentiation and allows us to transform the integral of a product into a potentially simpler form.

• Formula:$\int u \, dv = uv - \int v \, du$

  Here, u and v are differentiable functions of x.

Key Steps for Using Integration by Parts

1. Identify parts of the integrand to assign as u and dv.

2. Differentiate u to find du, and integrate dv to find v.

3. Substitute into the integration by parts formula.

4. Simplify and integrate any remaining terms.

Conditions for Integration by Parts

Integration by parts can be used only if the integrand satisfies the following conditions:

1. The integrand can be written as the product of two factors, $u$ and $du$.

2. It is possible to integrate $dv$ to find $v$ and to differentiate $u$to find $du$.

   Example 1: $\int x e^x \, dx$

   1. IDENTIFY: Let$u = x$, and $dv=e^{x}dx$

   2. DIFFERENTIATE: $du=dx$and $v=e^{x}$

   3. SUBSTITUTE: $=uv-\int v\,du$

      $u=x$

      $v=e^{x}$

      $du=$$dx$

      $=xe^{x}-\int e^{x}dx$

   4. SIMPLIFY & INTEGRATE: $xe^{x}-\int e^{x}dx$

      $=xe^{x}-e^{x}+C$

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

   1. IDENTIFY: parts of the integrand to assign as u and dv.

      Let $u=\ln x$(this is what we'll differentiate)

      and $dv=$$dx$(this is what we'll integrate)

      Note: Even though we only see "ln x" and not two obvious functions multiplied, we can think of it as (ln x)·(1), so dv = 1·dx

   2. DIFFERENTIATE: to find du, and integrate dv to find v.

      Differentiate u:

      u = ln x

      $du=\frac{1}{x}dx$

      Integrate dv:

      $v=x$

      $dv=$$dx$

   3. SUBSTITUTE: into the integration by parts formula.

      Substitute:

      $u=\ln x$

      $v=x$

      $du=\frac{1}{x}dx$

      Apply the formula:$\int u\,dv=uv-\int v\,du$

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

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

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

   4. SIMPLIFY & INTEGRATE: any remaining terms.

      Simplify:

      $x\cdot\frac{1}{x}dx$$=\int1dx=x$

      $=x\cdot\ln x-x+C$

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

   1. Identify parts of the integrand to assign as u and dv.

   2. Differentiate u to find du, and integrate dv to find v.

   3. Substitute into the integration by parts formula.

   4. Simplify and integrate any remaining terms.

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

   2. Differentiate and integrate as needed, possibly applying integration by parts more than once.

   3. Additional info: This example demonstrates repeated use of integration by parts for polynomials times exponentials.

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

   1. Let $u = \ln x$, $dv = x^2 dx$.

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

   3. Apply the formula:

      $\int x^2 \ln x dx = \frac{1}{3} x^3 \ln x - \int \frac{1}{3} x^3 \cdot \frac{1}{x} dx = \frac{1}{3} x^3 \ln x - \frac{1}{3} \int x^2 dx = \frac{1}{3} x^3 \ln x - \frac{1}{9} x^3 + C$

Summary Table: Integration by Parts Process

Step	Description
1. Choose $u$and $du$	Pick $u$ to differentiate easily, $dv$ to integrate easily
2. Differentiate $u$	Find $du$
3. Integrate $dv$	Find $v$
4. Substitute	Apply $\int u \, dv = uv - \int v \, du$
5. Simplify	Integrate remaining terms and combine results

Applications

• Integration by parts is widely used in calculus, physics, and engineering to solve integrals involving products of polynomials, exponentials, logarithms, and trigonometric functions.

• It is also essential for solving certain differential equations and for evaluating definite integrals.