##### Integration by Partial Fraction

If the function that need to be integrated is in the form of an algebraic fraction which is not easy to evaluate then you need to write the fraction into partial fraction to make it simpler for integration. You have to remember that this method is done only if the degree of the numerator is less than the degree of the denominator.

The method to write the fraction into partial fraction is by partial fraction decomposition.

### How to decompose an algebraic fraction into partial fraction?

We will have a rational expression on the form $\displaystyle f(x)=\frac{{P(x)}}{{Q(x)}}$ where both $\displaystyle P(x)$ and$\displaystyle Q(x)\ne 0$ are polynomials and the degree of  $\displaystyle P(x)$ is smaller than the degree of $\displaystyle Q(x)$.

After determining that it can turn into a partial fraction then we factor the denominator as completely as possible.

For each factor in the denominator we can use the table of partial fraction decomposition:

### Denominator Containing Linear factor

Example 1: Find the integral $\displaystyle \int{{\frac{{2x+11}}{{(x-1)(x+2)}}}}dx$

Firstly we see that the degree of the numerator is less than the degree of the denominator so we can turn it into a partial fraction.

The denominator is factored so we turn it into a partial fraction:

$\displaystyle \frac{{2x+11}}{{(x-1)(x+2)}}=\frac{A}{{(x-1)}}+\frac{B}{{(x+2)}}$

We find the value of$\displaystyle A$ and $\displaystyle B$ by multiplying both sides by$\displaystyle {(x-1)}$ and $\displaystyle {(x+2)}$:

$\displaystyle 2x+11=A(x+2)+B(x-1)$

Expand and collect the like terms:

$\displaystyle 2x+11=Ax+2A+Bx-B$

$\displaystyle 2x+11=(A+B)x+(2A-B)$

Create a system of equations with A and B by equating coefficients:

$\displaystyle \left\{ \begin{array}{l}A+B=2\\2A-B=11\end{array} \right.$

Solving the simultaneous equations we get:

$\displaystyle B=2-A$

$\displaystyle 2A-(2-A)=11$

$\displaystyle 2A-2+A=11$

$\displaystyle 3A=13$

$\displaystyle A=\frac{{13}}{3}$

$\displaystyle B=2-\frac{{13}}{3}=\frac{{6-13}}{3}=-\frac{7}{3}$

We found that $\displaystyle A=\frac{{13}}{3}$ and $\displaystyle B=-\frac{7}{3}$

So $\displaystyle \frac{{2x+11}}{{(x-1)(x+2)}}=\frac{{{}^{{13}}\!\!\diagup\!\!{}_{3}\;}}{{(x-1)}}+\frac{{-{}^{7}\!\!\diagup\!\!{}_{3}\;}}{{(x+2)}}$

Now we find the integral:

$\displaystyle \int{{\frac{{2x+11}}{{(x-1)(x+2)}}}}dx$$\displaystyle =\frac{{13}}{3}\int{{\frac{1}{{(x-1)}}dx}}-\frac{7}{3}\int{{\frac{1}{{(x+2)}}dx}} \displaystyle \int{{\frac{{2x+11}}{{(x-1)(x+2)}}}}dx$$\displaystyle =\frac{{13}}{3}\ln (x-1)-\frac{7}{3}\ln (x+2)+C$

Example 2: Find the integral $\displaystyle \int{{\frac{{3x}}{{6{{x}^{2}}+7x+2}}}}dx$?

Firstly we see that the degree of the numerator is less than the degree of the denominator so we can turn it into a partial fraction.

Now we factor the denominator:

$\displaystyle \frac{{3x}}{{6{{x}^{2}}+7x+2}}=\frac{{3x}}{{(2x+1)(3x+2)}}$

Then use the table to turn it into a partial fraction:

$\displaystyle \frac{{3x}}{{(2x+1)(3x+2)}}$$\displaystyle =\frac{A}{{(2x+1)}}+\frac{B}{{(3x+2)}} We find the value of \displaystyle A and \displaystyle B by multiplying both sides by \displaystyle {(2x+1)} and \displaystyle {(3x+2)}: \displaystyle 3x=A(3x+2)+B(2x+1) Expand and collect the like terms: \displaystyle 3x=3xA+2A+2xB+B \displaystyle 3x=(3A+2B)x+(2A+B) Create a system of equations with A and B by equating coefficients: \displaystyle \left\{ \begin{array}{l}3A+2B=3\\2A+B=0\end{array} \right. Solving the simultaneous equations we get: \displaystyle A=\frac{{-B}}{2} \displaystyle 3(\frac{{-B}}{2})+2B=3 \displaystyle \frac{{-3B}}{2}+\frac{{4B}}{2}=3 \displaystyle \frac{B}{2}=3 \displaystyle B=6 \displaystyle A=\frac{{-B}}{2}=\frac{{-6}}{2}=-3 We found that \displaystyle A=-3 and \displaystyle B=6 So \displaystyle \frac{{3x}}{{(2x+1)(3x+2)}}=\frac{{-3}}{{(2x+1)}}+\frac{6}{{(3x+2)}} Now we find the integral: \displaystyle \int{{\frac{{3x}}{{6{{x}^{2}}+7x+2}}}}dx$$\displaystyle =\int{{\frac{{-3}}{{(2x+1)}}dx+\int{{\frac{6}{{(3x+2)}}}}}}dx$

To find each integral we use the u-substitution method:

$\displaystyle \int{{\frac{{-3}}{{(2x+1)}}}}dx$

$\displaystyle u=2x+1$ than $\displaystyle du=2dx$

We multiply and divide with 2 to fix the coefficients and find the integral

$\displaystyle \int{{\frac{{-3}}{{(2x+1)}}}}dx$$\displaystyle =\frac{{-3}}{2}\int{{\frac{{du}}{u}}}$$\displaystyle =-\frac{3}{2}\ln u+C$

Then substitute back the u:$\displaystyle \int{{\frac{{-3}}{{(2x+1)}}}}dx=-\frac{3}{2}\ln (2x+1)+C$

We find the other integral with the same method

$\displaystyle \int{{\frac{6}{{(3x+2)}}}}dx=\int{{\frac{{2du}}{u}}}=2\ln u+C$

$\displaystyle \int{{\frac{6}{{(3x+2)}}}}dx=2\ln (3x+2)+C$

$\displaystyle \int{{\frac{{3x}}{{6{{x}^{2}}+7x+2}}}}dx$$\displaystyle =-\frac{3}{2}\ln (2x+1)+2\ln (3x+2)+C ### Denominator Containing Repeated Linear Factor Example 3: Find the integral \displaystyle \int{{\frac{{5{{x}^{2}}+1}}{{{{{(x+1)}}^{3}}}}}}dx? Firstly we see that the degree of the numerator is less than the degree of the denominator so we can turn it into a partial fraction. Then use the table to turn it into a partial fraction: \displaystyle \frac{{5{{x}^{2}}+1}}{{{{{(x+1)}}^{3}}}}$$\displaystyle =\frac{A}{{(x+1)}}+\frac{B}{{{{{(x+1)}}^{2}}}}+\frac{C}{{{{{(x+1)}}^{3}}}}$

We multiply the coefficients with the factor it’s missing:

$\displaystyle 5{{x}^{2}}+1=A{{(x+1)}^{2}}+B(x+1)+C$

Expand and collect the like terms:

$\displaystyle 5{{x}^{2}}+1$$\displaystyle =A{{x}^{2}}+2Ax+A+Bx+B+C \displaystyle 5{{x}^{2}}+1$$\displaystyle =A{{x}^{2}}+(2A+B)x+A+B+C$

Create a system of equations with A, B and C by equating coefficients:

$\displaystyle \left\{ \begin{array}{l}A=5\\2A+B=0\\A+B+C=1\end{array} \right.$

Solving the simultaneous equations we get:

$\displaystyle A=5$, $\displaystyle B=-10$, $\displaystyle C=6$

So$\displaystyle \frac{{5{{x}^{2}}+1}}{{{{{(x+1)}}^{3}}}}=\frac{5}{{(x+1)}}+\frac{{-10}}{{{{{(x+1)}}^{2}}}}+\frac{6}{{{{{(x+1)}}^{3}}}}$

Now we find the integral, for the second and third we use u-substitution method and power rule of integration:

$\displaystyle \int{{\frac{5}{{(x+1)}}dx}}=5\ln (x+1)+{{C}_{1}}$

$\displaystyle \int{{\frac{{10}}{{{{{(x+1)}}^{2}}}}dx}}$$\displaystyle =10\frac{{{{{\left( {x+1} \right)}}^{{-2+1}}}}}{{-2+1}}+{{C}_{2}}$$\displaystyle =\frac{{-10}}{{(x+1)}}+{{C}_{2}}$

$\displaystyle \int{{\frac{6}{{{{{(x+1)}}^{3}}}}dx}}$$\displaystyle =6\frac{{{{{(x+1)}}^{{-3+1}}}}}{{-3+1}}+{{C}_{3}}$$\displaystyle =\frac{{-3}}{{{{{(x+1)}}^{2}}}}+{{C}_{3}}$

The answer is: $\displaystyle \int{{\frac{{5{{x}^{2}}+1}}{{{{{(x+1)}}^{3}}}}dx}}=5\ln (x+1)-\frac{{10}}{{(x+1)}}-\frac{3}{{{{{(x+1)}}^{2}}}}+C$

### Denominator Containing a Quadratic Factor

Example 4: Find the integral $\displaystyle \int{{\frac{{3x-1}}{{x({{x}^{2}}+1)}}dx}}$?

Firstly we see that the degree of the numerator is less than the degree of the denominator so we can turn it into a partial fraction.

Then use the table o turn it into a partial fraction:

$\displaystyle \frac{{3x-1}}{{x({{x}^{2}}+1)}}=\frac{A}{x}+\frac{{Bx+C}}{{{{x}^{2}}+1}}$

We multiply the coefficients with the factor it’s missing:

$\displaystyle 3x-1=A({{x}^{2}}+1)+(Bx+C)x$

Expand and collect the like terms:

$\displaystyle 3x-1=A{{x}^{2}}+A+B{{x}^{2}}+Cx$

$\displaystyle 3x-1=(A+B){{x}^{2}}+Cx+A$

Create a system of equations with A, B and C by equating coefficients:

$\displaystyle \left\{ \begin{array}{l}A+B=0\\C=3\\A=-1\end{array} \right.$

Then find the coefficients:

$\displaystyle A=-1$,$\displaystyle B=1$ and $\displaystyle C=3$

Now we find the integral:

$\displaystyle \int{{\frac{{3x-1}}{{x({{x}^{2}}+1)}}dx}}=\int{{\frac{{-1}}{x}dx}}+\int{{\frac{{x+3}}{{{{x}^{2}}+1}}}}dx$

The first integral is easy, but the second one needs a little bit of work

$\displaystyle \int{{\frac{{-1}}{x}dx}}=-\ln x+{{C}_{1}}$

The second interval we are going to divide it and then apply the sum rule:

$\displaystyle \int{{\frac{{x+3}}{{{{x}^{2}}+1}}}}dx=\int{{\frac{x}{{{{x}^{2}}+1}}}}dx+\int{{\frac{3}{{{{x}^{2}}+1}}}}dx$

For the first one we are going to use the u-substitution method and the second one is a table integral.

$\displaystyle \int{{\frac{x}{{{{x}^{2}}+1}}}}dx$

$\displaystyle u={{x}^{2}}+1$ than  $\displaystyle du=2xdx$

We fix the coefficients by multiplying and dividing with 2.

$\displaystyle \int{{\frac{x}{{{{x}^{2}}+1}}}}dx=\int{{\frac{2}{2}}}(\frac{x}{{{{x}^{2}}+1}})dx=\frac{1}{2}\int{{\frac{{du}}{u}}}$

$\displaystyle \frac{1}{2}\int{{\frac{{du}}{u}}}=\frac{1}{2}\ln u+{{C}_{2}}$

$\displaystyle \int{{\frac{x}{{{{x}^{2}}+1}}}}dx=\frac{1}{2}\ln ({{x}^{2}}+1)+{{C}_{2}}$

The second integral: $\displaystyle \int{{\frac{3}{{{{x}^{2}}+1}}}}dx=3\arctan ({{x}^{2}}+3)+{{C}_{3}}$

$\displaystyle \int{{\frac{{3x-1}}{{x({{x}^{2}}+1)}}dx}}=-\ln x+\frac{1}{2}\ln ({{x}^{2}}+1)+3\arctan ({{x}^{2}}+3)+C$

Note! This type of integral seems difficult but if you follow the steps one by one and observe each integral carefully you will find the correct answer.

Example 5Find the integral $\displaystyle \int{{\frac{{2{{x}^{2}}+3x-2}}{{({{x}^{2}}+x-1)(x-2)}}dx}}$?

Firstly we see that the degree of the numerator is less than the degree of the denominator so we can turn it into a partial fraction.

Then use the table to turn it into a partial fraction:

$\displaystyle \frac{{2{{x}^{2}}+3x-2}}{{({{x}^{2}}+x-1)(x-2)}}$$\displaystyle =\frac{{Ax+B}}{{{{x}^{2}}+x-1}}+\frac{C}{{x-2}} We multiply the coefficients with the factor it’s missing: \displaystyle 2{{x}^{2}}+3x-2=(Ax+B)(x-2)+C({{x}^{2}}+x-1) Expand and collect the like terms: \displaystyle 2{{x}^{2}}+3x-2$$\displaystyle =A{{x}^{2}}-2Ax+Bx-2B+C{{x}^{2}}+Cx-C$

$\displaystyle 2{{x}^{2}}+3x-2$$\displaystyle =(A+C){{x}^{2}}+(B-2A+C)x-2B-C Create a system of equations with A, B and C by equating coefficients: \displaystyle \left\{ \begin{array}{l}A+C=2\\B-2A+C=3\\-2B-C=-2\end{array} \right. Solving simultaneous equations we get: Using substitution method we find that: \displaystyle A=-\frac{2}{5}, \displaystyle B=-\frac{1}{5} and \displaystyle C=\frac{{12}}{5} Now we find the integral: \displaystyle \int{{\frac{{2{{x}^{2}}+3x-2}}{{({{x}^{2}}+x-1)(x-2)}}dx}} After observing both integrals we see that we can solve them by using u-substitution method. The first integral: \displaystyle \int{{\frac{{{}^{{12}}\!\!\diagup\!\!{}_{5}\;}}{{x-2}}dx}}=\frac{{12}}{5}\ln (x-2)+{{C}_{1}} Let’s find the second integral: \displaystyle -\frac{1}{5}\int{{\frac{{2x+1}}{{{{x}^{2}}+x-1}}}}dx=-\frac{1}{5}\int{{\frac{{du}}{u}}}=-\frac{1}{5}\ln u+{{C}_{2}} \displaystyle -\frac{1}{5}\int{{\frac{{2x+1}}{{{{x}^{2}}+x-1}}}}dx=-\frac{1}{5}\ln ({{x}^{2}}+x-1)+{{C}_{2}} The answer is: \displaystyle \int{{\frac{{2{{x}^{2}}+3x-2}}{{({{x}^{2}}+x-1)(x-2)}}dx}}$$\displaystyle =-\frac{1}{5}\ln ({{x}^{2}}+x-1)+\frac{{12}}{5}\ln (x-2)+C$