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Integration of u/v dx

Last Updated : 23 Jul, 2025
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Integration is a way to add up small pieces to find a total. Imagine you're tracking how fast you're going while driving. If you know your speed at different times, integration helps you figure out how far you've traveled in total.

In math, integration is the opposite of finding the rate of change (called differentiation). While differentiation tells us how fast something is changing, integration helps us add up those changes to find the overall result.

Techniques for Integrating u/v

To integrate a function of the form u/v​, where u and v are functions of x, you can apply integration by parts or substitution, but often the best method is integration by parts when the function fits that pattern. The general form for integration by parts is based on the product rule of differentiation.

However, integration by parts doesn't directly apply to u/v​. Instead, you should first simplify the expression using a technique like partial fraction decomposition or substitution, or use special methods for rational functions.

Integration by Parts

Sometimes, if u/v​ is a product of two functions, you can rewrite it and apply integration by parts, where:

\int u \frac{dv}{dx} = uv - \int v \frac{du}{dx}

Let's consider an example for better understanding.

Example: Consider the integral: Integrate ln(x)/x dx.

Solution:

Given: I = ln(x)/x dx

Let u = ln(x) (which simplifies when differentiated), then dv = (1/x) dx.

  • Differentiate u: du = (1/x) dx.
  • Integrate dv: v = ln(x).

I = ∫ln(x)/x dx = u v - ∫v du,
I = ln(x) × ln(x) - ∫ ln(x) × (1/x) dx
I = [ln(x)]2- I
⇒ 2I = [ln(x)]2
⇒ I = [ln(x)]2/2 + C

Partial Fraction Decomposition

If u(x) and v(x) are both polynomials, you can break the fraction into simpler terms using partial fraction decomposition.

  • Rewrite u(x)/v(x)​ as a sum of simpler fractions.
  • Integrate each fraction separately.

Example: To integrate \frac{2x + 3}{x^2 + x}​, decompose it into:

\frac{2x + 3}{x^2 + x} = \frac{A}{x} + \frac{B}{x + 1}

Find A and B and integrate each part.

Read More about Partial Fraction.

Integration by Substitution

When v(x) is a function whose derivative is related to u(x), you can use substitution to simplify the integral.

  • Let v(x) = t, so dv = v′(x)dx.
  • Replace the variable x in the integral to get a simpler form in terms of t.

Let's consider an example for better understanding.

Example: Consider the integral: Integrate x/(x2 + 1) dx.

Solution:

Given: I = x/(x2 + 1) dx

Let u = x2+ 1 ⇒ du = 2x dx, or equivalently, x dx = du/2.
Thus, I = 1/(2u) du
I = (1/2) × ln |u| + C = (1/2) × ln |x2 + 1| + C.

Therefore, the integral of x/(x^2 + 1) dx is (1/2) * ln |x^2 + 1| + C.

Read More about Substitution Method.

Some Special Cases

Some special cases of integration of u/v form of functions are:

  • ∫ex/x dx is a standard form, and the result is known as the Exponential Integral. The integral does not have a simple expression and is denoted as Ei(x) + C.
  • For forms like \frac{1}{x^2 \pm a^2}, use trigonometric substitution.

Solved Problems on Integration of u/v dx

Problem 1: Integrate 2x/(x2 + 1) dx.

Solution:

Given: I = 2x/(x2 + 1) dx
Let u = x2 + 1, then du = 2x dx
∫(1/u) du = ln |u| + C = ln |x2 + 1| + C.

Problem 2: Integrate (1 + sin(x))/cos(x) dx

Solution:

Break into two parts: integral (1/cos(x)) dx + integral (sin(x)/cos(x)) dx.

This results in: ln |sec(x) + tan(x)| + ln |sec(x)| + C.

Problem 3: Integrate (x2 + 1)/(x2) dx.

Solution:

I = ∫(1 + 1/x2) dx = ∫dx + ∫(1/x2) dx.
I = x - 1/x + C.

Worksheet on Integration of u/v dx

Worksheet-on-Integration-of-u-by-v-dx

You can download this free worksheet on Integration of u/v dx with answers from below:

Download Free Worksheet on Integration of u/v dx

Conclusion

In conclusion, integrating the expression u/v dx involves several techniques that help simplify and solve the integral. Whether using substitution to make the integral easier, partial fraction decomposition for rational functions, or integration by parts for products of functions, these methods allow us to break down complex integrals into simpler parts.

Read More,


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