Skip to content
Mom's Basement University

Review of Basic Integration and Substitution

What You’ll Learn

Section titled “What You’ll Learn”

This lesson reviews the integration tools from Calculus 1 to make sure your foundation is solid before we add new techniques.

The Concept

Section titled “The Concept”

Basic Integration Rules

Section titled “Basic Integration Rules”

Power rule:

∫xn dx=xn+1n+1+C(n≠−1)\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)

Special cases:

∫1x dx=ln⁡∣x∣+C\int \frac{1}{x} \, dx = \ln|x| + C ∫ex dx=ex+C\int e^x \, dx = e^x + C

Trig:

∫sin⁡x dx=−cos⁡x+C∫cos⁡x dx=sin⁡x+C\int \sin x \, dx = -\cos x + C \qquad \int \cos x \, dx = \sin x + C

Constants factor out, and sums/differences split term by term. These rules handle any polynomial, basic trig, and exponential integral directly.

U-Substitution

Section titled “U-Substitution”

When the integrand contains a composite function and the derivative of the inside is present (or close to it), substitution simplifies the integral.

  1. Choose u = the inside function
  2. Compute du
  3. Rewrite entirely in terms of u and du
  4. Integrate
  5. Substitute back to x

For a general exponential:

∫ekx dx=1kekx+C\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C

This comes from substituting u = kx, du = k dx.

Worked Example

Section titled “Worked Example”

Example 1: Polynomial

∫(4x3−6x+7) dx=x4−3x2+7x+C\int (4x^3 - 6x + 7)\,dx = x^4 - 3x^2 + 7x + C

Each term uses the power rule independently.

Example 2: Basic substitution

Evaluate:

∫2x(x2+5)3 dx\int 2x(x^2 + 5)^3\,dx

Let u = x² + 5, du = 2x dx.

∫u3 du=u44+C=(x2+5)44+C\int u^3\,du = \frac{u^4}{4} + C = \frac{(x^2 + 5)^4}{4} + C

Example 3: Trig with substitution

Evaluate:

∫sin⁡(4x) dx\int \sin(4x)\,dx

Let u = 4x, du = 4 dx, so dx = du/4.

14∫sin⁡u du=−14cos⁡u+C=−14cos⁡(4x)+C\frac{1}{4}\int \sin u\,du = -\frac{1}{4}\cos u + C = -\frac{1}{4}\cos(4x) + C

Example 4: Square root substitution

Evaluate:

∫xx2−1 dx\int x\sqrt{x^2 - 1}\,dx

Let u = x² - 1, du = 2x dx, so x dx = du/2.

12∫u du=12⋅23u3/2+C=13(x2−1)3/2+C\frac{1}{2}\int \sqrt{u}\,du = \frac{1}{2}\cdot\frac{2}{3}u^{3/2} + C = \frac{1}{3}(x^2 - 1)^{3/2} + C

Real-World Application

Section titled “Real-World Application”

These are the building blocks for everything in Calculus 2:

  • Every advanced technique (integration by parts, partial fractions, trig substitution) eventually reduces to basic rules
  • Physics problems start with setting up the integral, then you need these rules to evaluate it
  • U-substitution is the most frequently used technique in all of calculus

Quiz

Section titled “Quiz”
The antiderivative of x⁵ is:
U-substitution is the reverse of:
The integral of sin(3x) dx equals:
After integrating with respect to u, you must:
The integral of e to the 5x dx is: