Calculus

# Partial Integration Explained

## Rachel McLean

Subject Matter Expert

In this article, we’ll discuss how to integrate partial derivatives using partial integration. To refresh our knowledge, we’ll also review the definition of integration and partial derivatives.

## What Is Partial Integration?

To understand partial integration, you need to know first the definition of integration. Integration is the process of finding a function’s antiderivative function. You can think about integration as the reverse operation of differentiation.

When you integrate some function $f(x)$, you find its antiderivative function. We often denoted this function as $F(x)$, and this process is called taking the indefinite integral. Using this function, a definite integral ‌calculates the area underneath the curve of $f(x)$ on a specific interval.

The notation for integrating $f(x)$ looks like this:

$\int f(x)\,dx = F(x) + C$

Taking the derivative of $F(x)$ outputs back $f(x)$, which is why $F(x)$ is called the antiderivative function. The fundamental theorem of calculus assures this relationship between integration and differentiation. In other words, if $\int f(x)\,dx = F(x) + C$, then $F’(x) = f(x)$. We can also say that $\int f’(x)\,dx = f(x) + C$.

But what happens if we are integrating a partial derivative?

As a reminder, we use partial differentiation to differentiate a function of two or more variables. Partial derivatives measure the rate of change of a multivariable function when ​​we let just one of its variables change, while the rest are held constant.

For example, consider the function $f(x, y) = xy + x^2y$, which is a function of two variables. To find $\frac{\partial{f}}{\partial{x}}$, we treat $y$ as a constant, and then differentiate the function normally. To find $\frac{\partial{f}}{\partial{y}}$, we treat $x$ as a constant, and then differentiate the function normally.

So for our function $f(x, y) = xy + x^2y$, the partial derivative with respect to $x$ is $\frac{\partial}{\partial{x}}f(x, y) = y + 2yx$. The partial derivative with respect to $y$ is $\frac{\partial}{\partial{y}}f(x, y) = x + x^2$.

A similar logic holds for partial integration. Suppose we take the integral of a partial derivative of $f(x, y)$. To integrate a partial derivative with respect to $x$, we would integrate with respect to $x$ while treating $y$ as a constant.

The resulting partial integration formula would look something like:

$\int \frac{\partial}{\partial x} f(x, y) \ dx = F(x, y) + C(y)$

To integrate a partial derivative with respect to $y$, we would integrate with respect to $y$ while treating $x$ as a constant.

This would look something like:

$\int \frac{\partial}{\partial y} f(x, y) \ dy = F(x, y) + C(x)$

Note that in this context, partial integration is different than integration by parts and integration by partial fractions. Here, partial integration outputs an integral of a partial derivative with respect to one variable of a multivariable function. By contrast, integration by parts is a technique used to integrate a product of functions, and partial fraction integration is a method used to integrate rational functions.

## How Do You Solve Partial Integration?

To use partial integration, we fix one variable as a constant. To integrate a partial derivative with respect to $x$, we must treat $y$ as a constant. To integrate a partial derivative with respect to $y$, we must treat $x$ as a constant.

Integrating a partial derivative with respect to one variable by treating the other variable as a constant looks like this:

$\int \frac{\partial}{\partial x} f(x, y) \ dx = F(x, y) + C(y)$

or

$\int \frac{\partial}{\partial y} f(x, y) \ dy = F(x, y) + C(x)$

Note that our integrand is a derivative. By the fundamental theorem of calculus, the integral of the derivative of a function should equal the original function itself. So, in the above cases, $F(x, y) + C(y) = f(x, y)$ and $F(x, y) + C(x) = f(x, y)$.

Dr. Hannah Fry is an associate professor at University College London. She discusses more about the fundamental theorem:

After we designate one variable as a constant, partial integration becomes nearly identical to normal integration. But, notice that $C(y)$ and $C(x)$ appear at the end of our antiderivative function, instead of the normal constant of integration $C$. Why is this?

When integrating a function of one variable, the constant $C$ always accompanies the antiderivative function, since the derivative of any constant evaluates to zero. Since $C$ can be any constant, $F(x) + C$ gives us a family of infinitely many functions whose derivative is $f(x, y)$.

Similarly, when taking the derivative of a function with respect to $x$, we treat the variable $y$ as a constant. As a result, the derivative of any function of $y$ evaluates to zero. Thus the expression $F(x, y) + C(y)$ represents a family of infinitely many functions whose derivative is $f(x, y)$.

Let’s walk through an example with our polynomial function $f(x, y) = xy + x^2y$, using its partial derivatives $\frac{\partial}{\partial{x}}f(x,y) = y + 2yx$ and $\frac{\partial}{\partial{y}}f(x,y) = x + x^2$.

First, let’s find the integral of the partial derivative with respect to $x$. So, we will treat $y$ as a constant when integrating $\frac{\partial}{\partial{x}}f(x,y) = y + 2yx$. To more easily imagine $y$ as a constant, one handy trick is to replace $y$with $c$ or $k$, which are two variables that are commonly used to represent constant values.

Let’s use this trick and replace $y$ with $k$. Then, we’ll use the power rule for integration. Remember to change $k$ back to $y$ at the end.

$\int \frac{\partial}{\partial x} f(x, y) \ dx = \int (y + 2yx) \ dx$
$= \int (k + 2kx) \ dx$
$= kx + \frac{2kx^2}{2}$
$= kx + kx^2$
$= xy + x^2y + C(y)$

Now, let’s find the integral of the partial derivative with respect to $y$. So, we will treat $x$ as a constant when integrating $\frac{\partial}{\partial{y}}f(x, y) = x + x^2$. To more easily imagine $x$ as a constant, we’ll replace $x$ with $k$.

Using the power rule for integration, we have:

$\int \frac{\partial}{\partial y} f(x, y) \ dy = \int (x + x^2) \ dy$
$= \int (k + k^2) \ dy$
$= ky + k^2y$
$= xy + x^2y + C(x)$

## Integration Rules

To quickly calculate antiderivative functions of partial derivatives, it’s helpful to memorize the list of integration rules below. Assume that $f$ and $g$ are continuous functions:

### Sum Rule

$\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx$

### Difference Rule

$\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx$

### Constant Multiplier Rule

$\int kf(x)\,dx = k\int f(x)\,dx$ for some constant $k$

### Power Rule

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for some real number $n$

### Constant Rule

$\int a\,dx = ax + C$ for some constant $a$

### Reciprocal Rules

$\int \frac{1}{x}\,dx = \int x^{-1}\,dx = \ln{|x|} + C$

$\int\frac{1}{ax+b}\,dx = \frac{1}{a} \ln{(ax+b)} + C$

### Exponential and Logarithmic Function Rules

$\int e^x\,dx = e^x + C$

$\int a^x\,dx = \frac{a^x}{\ln{(a)}} + C$, for any positive real number $a$

$\int \ln{(x)}\,dx = x\ln{(x)}-x + C$

### Trigonometric Function Rules

For these rules, assume that x is in radians. $\int \sin{x}\,dx = -\cos{x} + C$

$\int \cos{x}\,dx = \sin{x} + C$

$\int \sec ^2 x\,dx = \tan{x} + C$

$\int \csc ^2 x\,dx = -\cot{x} + C$

$\int \sec{x}\tan{x}\,dx = \sec{x} + C$

$\int \csc{x}\cot{x}\,dx = -\csc{x} + C$

$\int \frac{dx}{\sqrt{1-x^2}}=\sin ^{-1}x + C$

$\int \frac{-dx}{\sqrt{1-x^2}}=\cos ^{-1}x + C$

$\int \frac{dx}{1+x^2}=\tan ^{-1}x + C$

$\int \sin{(ax)}\,dx = \frac{-\cos{(ax)}}{a}+C$ for some real number $a$

$\int \cos{(ax)}\,dx = \frac{\sin{(ax)}}{a}+C$ for some real number $a$

### Absolute Value Rule

$\int |x|\,dx = \frac{x |x|}{2} + C$

Dr. Hannah Fry explains more about integration rules in this lecture clip:

## Partial Integration Examples

### Example 1

Suppose that $\frac{\partial}{\partial x} f(x, y) = xy^2 + \sin{(x)}$ is the partial derivative with respect to $y$ of some function $f(x, y)$. Find $f(x, y)$.

### Solution

We’ll integrate with respect to $y$, so we must treat $x$ as a constant. Then, using the sum rule, power rule, and trigonometry rules, we have:

$\int \frac{\partial}{\partial x} f(x, y) \ dy = \int (xy^2 + \sin{(x)}) \ dy$
$= \int (ky^2 + \sin{(k)}) \ dy$
$= \frac{ky^3}{3} + y\sin{(k)}$
$= \frac{xy^3}{3} + y\sin{(x)} + C(x)$

Thus $f(x, y) = \frac{xy^3}{3} + y\sin{(x)} + C(x)$. To check our answer, we can take the partial derivative of $f(x, y)$ with respect to $y$. By hand or with an integral calculator, the resulting partial derivative is $\frac{\partial}{\partial x} f(x, y) = xy^2 + \sin{(x)}$, which is the partial derivative we started with.

### Example 2

Suppose that $\frac{\partial}{\partial x} f(x, y) = ye^x + \frac{y}{x}$ is the partial derivative with respect to $x$ of some function $f(x, y)$. Find $f(x, y)$.

### Solution

We’ll integrate with respect to $x$, so we must treat $y$ as a constant. Then, using the sum rule, exponential rule, and logarithmic rule, we have:

$\int \frac{\partial}{\partial x} f(x, y) \ dx = \int (ye^x + \frac{y}{x}) \ dx$
$= \int (ke^x + \frac{k}{x}) \ dx$
$= ke^x + k\ln{(x)}$
$= ye^x + y\ln{(x)} + C(y)$

Thus $f(x, y) = ye^x + y\ln{(x)} + C(y)$.

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