Integration by parts is a powerful technique for integrating products of functions, which is based upon the product rule for differentiation. It allows us to transform the integral of a product into simpler integrals that we can evaluate more easily.

Fundamental Formula

Use for differentiable functions $u$ and $v$:

For definite integrals from $a$ to $b$:

Choose $u$ using LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential).

Example Problems

Example 1: Integrating $x \sin(2x)$

1. Identify $u$ and $dv$:

• $u = x$, $dv = \sin(2x) dx$.

2. Differentiate and Integrate:

• $du = dx$, $v = \int \sin(2x) dx = -\frac{1}{2} \cos(2x)$.

3. Apply the Formula:

• $\int x \sin(2x) dx = x \left(-\frac{1}{2} \cos(2x)\right) - \int -\frac{1}{2} \cos(2x) dx$.

4. Simplify:

• Result: $-\frac{1}{2} x \cos(2x) + \frac{1}{4} \sin(2x) + C$.

Example 2: Integrating $x^2 e^{-x}$

1. Identify $u$ and $dv$:

• $u = x^2$, $dv = e^{-x} dx$.

2. Differentiate and Integrate:

• $du = 2x dx$, $v = \int e^{-x} dx = -e^{-x}$.

3. Apply the Formula:

• $\int x^2 e^{-x} dx = -x^2 e^{-x} - \int -2x e^{-x} dx$.

4. Simplify:

• Result: $(-2 - 2x - x^2) e^{-x} + C$.

Example 3: Integrating $\ln(x)$

1. Identify $u$ and $dv$:

• $u = \ln(x)$, $dv = dx$.

2. Differentiate and Integrate:

• $du = \frac{1}{x} dx$, $v = x$.

3. Apply the Formula:

• $\int \ln(x) dx = x \ln(x) - \int x \cdot \frac{1}{x} dx$.

4. Simplify:

• Result: $x \ln(x) - x + C$.

Example 4: Integrating $x \tan^{-1}(x)$

1. Identify $u$ and $dv$:

• $u = \tan^{-1}(x)$, $dv = x dx$.

2. Differentiate and Integrate:

• $du = \frac{1}{1 + x^2} dx$, $v = \frac{x^2}{2}$.

3. Apply the Formula:

• $\int x \tan^{-1}(x) dx = \frac{x^2}{2} \tan^{-1}(x) - \int \frac{x^2}{2} \cdot \frac{1}{1 + x^2} dx.$

4. Simplify:

• Result: $\frac{-x + (1 + x^2) \tan^{-1}(x)}{2} + C$.

Worked Example:

For $\int_1^4 \frac{\ln x}{\sqrt{x}} \,dx$:

Solution:

1. Rewrite as $\int_1^4 \ln x \cdot x^{-1/2} \,dx$.

2. Choose $u = \ln x$, $dv = x^{-1/2}dx$.

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

4. Apply formula: $[2x^{1/2}\ln x]_1^4 - \int_1^4 2 \cdot x^{-1/2} \,dx$.

5. Evaluate to get $4\ln 4 - 4$.