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Fraction Integration Techniques

12.07.2022 • 10 min read

Rachel McLean

Subject Matter Expert

Learn how to integrate a fraction using partial fractions, trigonometric identities, and substitution. There’s also solved examples and FAQs.

In This Article

  1. What Are Partial Fractions?

  2. 3 More Ways To Integrate a Fraction

  3. Integration Examples

  4. Frequently Asked Questions

  5. Learn More About Integration & Other Calculus Topics

What Are Partial Fractions?

Integrating fractions might seem daunting at first, but don’t fret! There are several handy integration techniques that students can use to make integrating fractions easier.

First, it’s good practice to memorize these reciprocal rules for integration. These rules help us to find the integral of partial fractions by using logarithms.

1xdx=x1dx=lnx+C\int \frac{1}{x}\, dx = \int x^{-1}dx = \ln{|x|} + C
1ax+bdx=1aln(ax+b)+C\int \frac{1}{ax+b}\, dx = \frac{1}{a} \ln{(ax+b)} + C

Partial fraction decomposition is one technique used to integrate fractions. In particular, we use the method of partial fractions to integrate proper rational functions.

We use partial fractions to decompose a rational function into simpler rational functions. In the same way, we can rewrite the rational expression 310 \frac{3}{10} as 1215 \frac{1}{2} - \frac{1}{5}. A rational function is a fraction whose numerator and denominator are both polynomials.

A proper rational function is a rational function where the degree of the numerator is smaller than the degree of the denominator. The degree of a polynomial is the largest exponent of the variable in the polynomial. If the degree of the numerator is larger than or equal to the degree of the denominator, then the function is called an improper function.

Proper functions:
xx23,x(x+1)(x+2)\frac{x}{x^2-3}\, , \frac{x}{(x+1)(x+2)}
Improper functions:
xx+2,,x51x41\frac{x}{x+2}\\, , \frac{x^5-1}{x^4-1}

When the denominator of a proper rational function is factorable, we can integrate by using partial fractions. This process involves splitting the integrand into several simpler rational functions with unknown numerators. Then, we can solve for the new unknown variables and integrate using the reciprocal rules for integrals.

  • If the rational fraction is in the form N(x)(ax+b)(cx+d)\frac{N(x)}{(ax+b)(cx+d)}, then the denominator is made up of linear factors, and so the form of partial fractions is Aax+b+Bcx+d\frac{A}{ax+b} + \frac{B}{cx+d}.

  • If the rational fraction is in the form N(x)(ax+b)2\frac{N(x)}{(ax+b)^2}, then the denominator is made up of repeated linear factors, and so the form of partial fractions is Aax+b+B(ax+b)2\frac{A}{ax+b} + \frac{B}{(ax+b)^2}.

  • If the rational fraction is in the form N(x)(ax+b)(cx+d)2\frac{N(x)}{(ax+b)(cx+d)^2}, then the denominator is made up of linear factors and repeated linear factors, and so the form of partial fractions is Aax+b+Bcx+d+C(cx+d)2\frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}.

  • If the rational fraction is in the form N(x)(ax+b)(x2+c2)\frac{N(x)}{(ax+b)(x^2+c^2)}, then the denominator is made up of linear factors and irreducible quadratic factors, and so the form of partial fractions is Aax+b+Bx+Cx2+c2\frac{A}{ax+b} + \frac{Bx+C}{x^2+c^2}

How To Integrate by Partial Fractions

Let’s break down the process further and detail each step involved in integration by partial fractions:

1. Factor the denominator of the function

If the function is an improper fraction, then use polynomial long division to rewrite the function as the sum of a proper fraction and a polynomial).

2. Decompose the function

Use the factored denominator to decompose the rational function into the sum of several simpler rational functions. To do this, use each factored term of the denominator as the denominator of a smaller fraction, and assign an unknown variable as the numerator of each smaller fraction.

3. Multiply both sides by the common denominator

After multiplying both sides by the common denominator, distribute and combine like terms.

4. Set coefficients equal

Set up one equation that equates the constant terms of the original function’s numerator with the constant terms in your new equation’s numerator. Set up a second equation that equates the “x” terms of the original function’s numerator with the “x” terms in your new equation’s numerator. Continue as necessary.

5. Solve the system

Solve for the unknown variables using a system of equations.

6. Plug in and integrate

Plug the solved variables into the decomposed fraction in Step 2. Integrate using the reciprocal rules to finish the integration process.

This process is easiest to understand with an example. Be sure to walk through the example covered in the later section.

3 More Ways To Integrate a Fraction

We’ll review three more methods used to integrate fractions.

1. Rewriting

Often, simply rewriting the fraction can reveal an easier way to integrate. It’s good practice to look closely at the function to determine if you can cleverly rewrite it before proceeding with a more complex method. Rewriting a fraction might include:

  • Simplifying the fraction by multiplying the numerator and denominator by the same number

  • Simplifying the fraction by dividing the numerator and denominator by the same number

  • Using negative powers in the numerator

  • Using polynomial long division to rewrite an improper fraction as the sum of a proper fraction and a polynomial

2. Trigonometric Identities

If the fraction consists of trigonometric functions, it can be helpful to use trigonometric identities to change the fraction’s form. Here are some trigonometric identities to know:

Pythagorean Identities

  • sin2x+cos2x=1\sin^2x + \cos^2x = 1

  • sec2xtan2x=1\sec^2 x - \tan^2 x = 1

  • csc2xcot2x=1\csc^2 x - \cot^2 x = 1

Quotient and Reciprocal Identities

  • tanx=sinxcosx\tan x = \frac{\sin x}{\cos x}

  • cotx=cosxsinx\cot x = \frac{\cos x}{\sin x}

  • sinx=1cscx\sin x = \frac{1}{\csc x}

  • cosx=1secx\cos x = \frac{1}{\sec x}

  • tanx=1cotx\tan x = \frac{1}{\cot x}

  • cscx=1sinx\csc x = \frac{1}{\sin x}

  • secx=1cosx\sec x = \frac{1}{\cos x}

  • cotx=1tanx\cot x = \frac{1}{\tan x}

Double Angle Identities

  • sin(2x)=2sinxcosx\sin{(2x)} = 2 \sin x \cos x

  • cos(2x)=cos2xsin2x=2cos2x1=12sin2x\cos{(2x)} = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x

  • tan(2x)=2tanx1tan2x\tan{(2x)} = \frac{2 \tan x}{1 - \tan^2 x}

3. Trigonometric Substitution

If the denominator of a function is not factorable, try using trigonometric substitution to integrate the fraction. In some cases, it might be necessary to complete the square of the denominator.

Here are the substitutions used in trigonometric substitution:

  • If you see \int \sqrt{a^2 - x^2} \, \dx , use x = a\sin{(\theta)}

  • If you see \int \sqrt{a^2 + x^2} \, \dx , use x=a\tan{(\theta)}

  • If you see \int \sqrt{x^2 - a^2} \, \dx , use x = a\sec{(\theta)}

Integration Examples

Example of Integrating by Partial Fractions

In the following partial fractions example, we’ll calculate 2x2x2+9x+4dx\int \frac{2x}{2x^2+9x+4} dx.

We’ll follow the steps detailed in the first section.

1. Factor the denominator of the function.

2x2x2+9x+4=2x(2x+1)(x+4)\frac{2x}{2x^2+9x+4} = \frac{2x}{(2x+1)(x+4)}

2. Decompose the function.

We’ll split the integrand into two smaller rational functions, using unknown numerators. We want to solve for AA and BB.

2x(2x+1)(x+4)=A2x+1+Bx+4\frac{2x}{(2x+1)(x+4)} = \frac{A}{2x+1} + \frac{B}{x+4}

3. Multiply both sides by the common denominator.

In this case, we multiply the left side and right side by (2x+1)(x+4)(2x+1)(x+4). This leaves us with:

2x=A(x+4)+B(2x+1)2x = A(x+4)+B(2x+1)
2x=Ax+4A+2Bx+B2x = Ax+4A+2Bx+B

4. Set coefficients equal.

Now, we collect the “x” terms together and factor. We also collect the constant terms together. See how this structure mirrors the structure of the numerator of our original integrand, 2x+02x+0?

2x+0=(Ax+2Bx)+(4A+B)2x+0 = (Ax+2Bx)+(4A+B)
2x+0=(A+2B)x+(4A+B)2x+0 = (A+2B)x +(4A+B)

5. Solve the system.

The next step is to set up a system of equations to solve for the unknowns, AA and BB. We’ll set up two equations.

First, we equate the “x” terms on each side of the equation.

2x=(A+2B)x2x = (A + 2B)x
2=A+2B2 = A + 2B

Next, we equate the constant terms on each side of the equation.

0=4A+B0 = 4A + B

We can now solve for AA and BB.

Solving the system of equations with 2=A+2B2 = A + 2B and 0=4A+B0 = 4A + B gives us A=27A=\frac{-2}{7} and B=87B=\frac{8}{7}.

6. Plug in and integrate.

Now that we’ve solved for AA and BB, we can plug these back into our function and calculate our integral using the reciprocal rules.

2x2x2+9x+4dx=(27(2x+1)+87(x+4))dx\int \frac{2x}{2x^2+9x+4} dx = \int(\frac{-2}{7(2x+1)}+\frac{8}{7(x+4)})\,dx
=2712x+1dx+871x+4dx=\frac{-2}{7}\int \frac{1}{2x+1}dx + \frac{8}{7}\int \frac{1}{x+4}\,dx
=2ln(2x+1)27+8ln(x+4)7+C=\frac{-2 \ln{(2x+1)}}{2 \cdot 7} + \frac{8 \ln{(x+4)}}{7} + C
=87ln(x+4)17ln(2x+1)+C=\frac{8}{7} \ln{(x+4)} -\frac{1}{7}\ln{(2x+1)} + C

Our final answer is 2x2x2+9x+4dx=87ln(x+4)17ln(2x+1)+C\int \frac{2x}{2x^2+9x+4} dx =\frac{8}{7} \ln{(x+4)} -\frac{1}{7}\ln{(2x+1)} + C.

Example of Integrating Fractions by Rewriting

Calculate the following integral x3x+7x2dx\int \frac{x^3- x + 7}{x^2}\, dx by rewriting the quotient.

Let’s try dividing the numerator and denominator by x2x^2. After doing this, the terms of the integral take a more familiar form we can solve using the power rule (xndx=xn+1n+1+C\int x^n\, dx = \frac{x^{n+1}}{n+1} + C) and inverse rule (1xdx=x1dx=lnx+C\int \frac{1}{x}\, dx = \int x^{-1}\, dx = \ln{|x|} + C).

x3x+7x2dx=(xx1+7x2)dx\int \frac{x^3 - x + 7}{x^2}\, dx = \int (x - x^{-1} + 7x^{-2})\, dx
=x22ln(x)7x1= \frac{x^2}{2} - \ln{(x)} - 7x^{-1}
=x227xln(x)+C= \frac{x^2}{2} - \frac{7}{x} - \ln{(x)} + C

Example of Integrating Using Trigonometric Identities

Calculate the following integral sin(x)1sin2(x)dx\int \frac{\sin{(x)}}{1-\sin^2{(x)}} \, dx using trigonometric identities.

Recall the trigonometric identity sin2(x)+cos2(x)=1\sin^2 {(x)} + \cos^2{(x)} = 1. We can rewrite this identity to say cos2(x)=1sin2(x)\cos^2{(x)} = 1 - \sin^2 {(x)}:

sin(x)1sin2(x)dx=sin(x)cos2(x)dx\int \frac{\sin{(x)}}{1-\sin^2{(x)}} \, dx = \int \frac{\sin{(x)}}{\cos^2{(x)}} \, dx

We’ll use our rewriting skills to transform the integrand further. Let’s break the fraction into the product of two simpler fractions.

sin(x)cos2(x)dx=sin(x)cos(x)1cos(x)dx \int \frac{\sin{(x)}}{\cos^2{(x)}} \, dx = \int \frac{\sin{(x)}}{\cos{(x)}} \cdot \frac {1}{\cos{(x)}} \, dx

We can use another trigonometric identity here. Recall that tanx=sinxcosx\tan x = \frac{\sin x}{\cos x} .

sin(x)cos(x)1cos(x)dx=tan(x)sec(x)dx \int \frac{\sin{(x)}}{\cos{(x)}} \cdot \frac {1}{\cos{(x)}} \, dx = \int \tan{(x)} \sec{(x)} \, dx

Our derivative formulas tell us that ddx(sec(x))=sec(x)tan(x) \frac{d}{dx} (\sec{(x)}) = \sec{(x)}\tan{(x)} . So, our final answer is:

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

Example of Integrating Using Trigonometric Substitution

Calculate the following integral x21x2dx \int \frac{x^2}{\sqrt{1-x^2}} \, dx using trigonometric substitution.

Note that our denominator is 1x2 \sqrt{1-x^2} . This is in the form a2x2 \sqrt{a^2 - x^2} where a=1 a = 1 . So, we can use the substitution x=asinθ=1sinθ=sinθx = a\sin{\theta} = 1\sin{\theta} = \sin{\theta} .

Since x=sinθ x = \sin{\theta} , we’ll also need to rewrite dx dx . In this case, we have dx=cosθdθ dx = \cos{\theta} \, d\theta . Now, using our substitutions, we have:

x21x2dx=sin2θ1sin2θcosθdθ\int \frac{x^2}{\sqrt{1-x^2}} \, dx = \int \frac{\sin^2{\theta}}{\sqrt{1-\sin^2{\theta}}} \cos{\theta} \, d\theta

Recall the Pythagorean identity that sin2x+cos2x=1 \sin^2x + \cos^2x = 1 . We can rewrite the denominator using this identity, and then simplify.

sin2θ1sin2θcosθdθ=sin2θcos2θcosθdθ \int \frac{\sin^2{\theta}}{\sqrt{1-\sin^2{\theta}}} \cos{\theta} \, d\theta = \int \frac{\sin^2{\theta}}{\sqrt{\cos^2{\theta}}} \cos{\theta} \, d\theta
=sin2θcosθcosθdθ= \int \frac{\sin^2{\theta}}{\cos{\theta}} \cos{\theta} \, d\theta
=sin2θdθ= \int \sin^2{\theta} \, d\theta

We can use the Double Angle identities to proceed. Recall that cos(2x)=12sin2x\cos{(2x)} = 1 - 2\sin^2 x . We can rewrite this to say sin2(x)=12(1cos(2x) \sin^2{(x)} = \frac{1}{2} (1 - \cos{(2x)} .

sin2(θ)dθ=12(1cos(2θ)dθ\int \sin^2{(\theta)} \, d\theta = \int {\frac{1}{2} (1 - \cos{(2\theta)} \, d\theta}
=12(θ12sin(2θ))+C= \frac{1}{2} (\theta - \frac{1}{2}\sin{(2\theta)} ) + C

We can also use the Double Angle identity sin(2x)=2sinxcosx\sin{(2x)} = 2 \sin x \cos x by rewriting it as 12sin(2x)=sinxcosx \frac{1}{2}\sin{(2x)} = \sin x \cos x :

12(θ12sin(2θ))+C=12(θsinθcosθ)+C \frac{1}{2} (\theta - \frac{1}{2}\sin{(2\theta)} ) + C = \frac{1}{2} (\theta - \sin{\theta} \cos{\theta}) + C

Now, all that’s left to do is rewrite our answer in terms of x x . Recall that our substitution was x=sin(θ)x = \sin{(\theta)} . Solving for θ \theta , we have θ=arcsinx \theta = \arcsin{x} .

You can draw this on a triangle and calculate all sides using the Pythagorean theorem, as below:

A triangle where you can calculate all sides using the Pythagorean theorem

Using this reference triangle, we finally can conclude that:

12(θ12sin(2θ))+C=12(arcsinxx1x2)+C \frac{1}{2} (\theta - \frac{1}{2}\sin{(2\theta)} ) + C = \frac{1}{2}(\arcsin{x} - x\sqrt{1-x^2}) + C

Frequently Asked Questions

How do you integrate improper fractions?

To integrate an improper fraction, perform polynomial long division to divide the numerator by the denominator. Then, rewrite the fraction as the sum of a proper fraction and a polynomial.

How do you integrate fractions with powers?

Any function in the form f(x)=axn f(x) = \frac{a}{x^n} can be written using a negative exponent in the form f(x)=axn f(x) = a \cdot x^{-n} . Then, we can use the power rule for integration to integrate the function, which states xndx=xn+1n+1+C \int x^ndx = \frac{x^{n+1}}{n+1} + C .

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