How to Integrate by Partial Fractions

Last Updated: September 3, 2020

When integrating functions involving polynomials in the denominator, partial fractions can be used to simplify integration. New students of calculus will find it handy to learn how to decompose functions into partial fractions not just for integration, but for more advanced studies as well.

Steps

1
Check to make sure that the fraction you are trying to integrate is proper. A proper fraction has a larger power in the denominator than in the numerator. If the power of the numerator is larger than or equal to the power of the denominator, it is improper and must be divided using long division.
- $\int {\frac {x^{{3}}-4x-10}{x^{{2}}-x-6}}{\mathrm {d}}x$
- In this example, the fraction is indeed improper because the power of the numerator, 3, is larger than the power of the denominator, 2. Therefore, long division must be used.
  - ${\frac {x^{{3}}-4x-10}{x^{{2}}-x-6}}=(x+1)+{\frac {3x-4}{x^{{2}}-x-6}}$
- The fraction is now proper. We can now split the integral into two parts. One of them containing the $x+1$ is easily evaluated, but we will evaluate at the end.
  - $\int (x+1){\mathrm {d}}x+\int {\frac {3x-4}{x^{{2}}-x-6}}{\mathrm {d}}x$
2
Factor the polynomials in the denominator.
- ${\frac {3x-4}{x^{{2}}-x-6}}={\frac {3x-4}{(x-3)(x+2)}}$
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3
Separate the fraction that you wish to decompose in to multiple fractions. The number of fractions in decomposition should equal the number of factors of $x.$ The numerators of these decomposed fractions should be represented with coefficients.
- ${\frac {3x-4}{(x-3)(x+2)}}={\frac {A}{x-3}}+{\frac {B}{x+2}}$
- If a factor of $x$ in the denominator has a power higher than 1, then the coefficients in the numerator should reflect this higher power. For example, a term in the denominator like $x^{{2}}+1$ that cannot be factored further can be represented with the term $Ax+B$ in the numerator.
  - ${\frac {Ax+B}{x^{{2}}+1}}+{\frac {C}{x-3}}$
- Roots of multiplicity more than 1 should be represented where both the root and its decreasing powers are written out, like so. An example of this below concerns a root of multiplicity 3. Notice that three fractions are written, where $(x+2)^{{3}},\ (x+2)^{{2}},$ and $(x+2)$ are all written out.
  - ${\frac {x-1}{(x+2)^{{3}}}}={\frac {A}{(x+2)^{{3}}}}+{\frac {B}{(x+2)^{{2}}}}+{\frac {C}{x+2}}$
- Let's return to the original example. We have now split up the fraction into its constituent parts. We can proceed in two different directions here. One method is to multiply everything out and solve a system of equations. Another, more efficient method is to recognize which terms go to zero and directly solve for the coefficients. This method will be outlined in the Substitution section.
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Method 1

Method 1 of 2:

System of Equations

1
Multiply both sides by the denominator of the original fraction in order to get rid of all denominators. Notice that right now, the right side is factored by coefficients.
- $3x-4=A(x+2)+B(x-3)$
2
Expand and factor. Instead of factoring by the coefficients $A$ and $B,$ we factor by powers of $x.$
- $3x-4=Ax+2A+Bx-3B=(A+B)x+(2A-3B)$
3
Set the coefficients equal on both sides. Because both sides are equal, that means that the coefficients of the $x$ terms are equal. We obtain a system of equations, where the number of equations depends on the degree of the denominator that you started out with.
- ${\begin{aligned}A+B&=3\\2A-3B&=-4\end{aligned}}$
4
Solve for all constants.
- $A=3-B$
- ${\begin{aligned}2A-3B&=-4\\2(3-B)-3B&=-4\\6-5B&=-4\\B&=2\end{aligned}}$
- $A=3-2=1$
5
Plug the coefficients into the decomposed fractions. Our integral is now ready to evaluate because we know the integral of ${\frac {1}{x}}.$
- $\int {\frac {3x-4}{x^{{2}}-x-6}}{\mathrm {d}}x=\int (x+1){\mathrm {d}}x+\int {\frac {1}{x-3}}{\mathrm {d}}x+\int {\frac {2}{x+2}}{\mathrm {d}}x$
6
Integrate . Although the u-subs are very easy to do, it is still recommended that you show all your work if you are not familiar with doing these types of integrals yet.
- $\int {\frac {3x-4}{x^{{2}}-x-6}}{\mathrm {d}}x={\frac {1}{2}}x^{{2}}+x+\ln |x-3|+2\ln |x+2|+c$
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Method 2

Method 2 of 2:

Substitution

1
Multiply both sides by $(x-3)$ and plug in $x=3$ . Notice that the term with $B$ in it goes to 0, but $A$ doesn't. Furthermore, multiplying everything by that factor makes sure that we don't get any division by 0 problems.
- ${\frac {3x-4}{(x-3)(x+2)}}={\frac {A}{x-3}}+{\frac {B}{x+2}}$
- ${\frac {3x-4}{x+2}}=A+{\frac {B(x-3)}{x+2}}$
- $A={\frac {3(3)-4}{(3)+2}}=1$
- This is a much more efficient method of solving for the coefficients as long as we think about which terms get sent to 0. Technically, when substituting these values, we are taking limits. But since our functions are easy to work with (polynomials), we don't need to worry about tricky discontinuity issues.
2
Multiply both sides by $(x+2)$ and plug in $x=-2$ . This solves for $B.$ Generally, we multiply by the factor and plug in the value of the root. That solves for the coefficient of the fraction whose denominator has that factor.
- ${\frac {3x-4}{x-3}}={\frac {A(x+2)}{x-3}}+B$
- $B={\frac {3(-2)-4}{(-2)-3}}=2$
3
Plug the coefficients into the decomposed fractions and integrate.
- $\int {\frac {3x-4}{x^{{2}}-x-6}}{\mathrm {d}}x={\frac {1}{2}}x^{{2}}+x+\ln |x-3|+2\ln |x+2|+c$
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Example 2: Repeated Roots

1
Consider the integral below. We use the previous example of a function whose factors in the denominator have multiplicity 3, but our numerator is a bit different.
- $\int {\frac {x^{{2}}-4x+9}{(x+2)^{{3}}}}{\mathrm {d}}x=\int \left({\frac {A}{(x+2)^{{3}}}}+{\frac {B}{(x+2)^{{2}}}}+{\frac {C}{x+2}}\right){\mathrm {d}}x$
2
Multiply both sides by $(x+2)^{{3}}$ . This immediately gets us $A$ if we plug in $x=-2.$
- $x^{{2}}-4x+9=A+B(x+2)+C(x+2)^{{2}}$
- $A=(-2)^{{2}}-4(-2)+9=21$
- However, we find that $B$ and $C$ cannot be obtained directly.
3
Differentiate once and plug in $x=-2$ to obtain $B$ .
- Let's start with where we are.
  - $x^{{2}}-4x+9=A+B(x+2)+C(x+2)^{{2}}$
- We see that the largest term containing a $B$ is a term with an $x.$ If we differentiate both sides, we know by the power rule that all that is left will be a constant. Meanwhile, $A$ goes away because that is already a constant. What does $C$ do? We can do the derivative for $C,$ or we can recognize that, whatever it is, there will still be an $(x+2)$ in the derivative, so after we plug in $x=-2,$ the term with $C$ vanishes as well.
  - $2x-4=B+2C(x+2)$
  - $B=2(-2)-4=-8$
4
Differentiate again and plug in $x=-2$ to obtain $C$ . Differentiating twice sends both $A$ and $B$ to 0, while only $C$ is left over. Be careful with the coefficient, though.
- $2=2C\to C=1$
5
Plug the coefficients into the decomposed fractions and integrate.
- ${\begin{aligned}\int {\frac {x^{{2}}-4x+9}{(x+2)^{{3}}}}{\mathrm {d}}x&=\int \left({\frac {21}{(x+2)^{{3}}}}+{\frac {-8}{(x+2)^{{2}}}}+{\frac {1}{x+2}}\right){\mathrm {d}}x\\&=-{\frac {21}{2}}{\frac {1}{(x+2)^{{2}}}}+{\frac {8}{x+2}}+\ln |x+2|+c\end{aligned}}$
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How do I get the values of multiple constants of integration by partial fractions?

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You only get a single constant of integration at the end of partial fractions because you can combine all constants into one larger constant.

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Do not forget the constant of integration $c$ if evaluating an indefinite integral!

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How to Integrate by Partial Fractions

Steps

System of Equations

Substitution

Example 2: Repeated Roots

Community Q&A

Tips

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