Question

How do you integrate $\int{\dfrac{2x+1}{\left( x-2 \right)\left( {{x}^{2}}+4 \right)}dx}$ using partial fractions?

Answer 1

This question belongs to an integration chapter. We have to integrate the given fraction using a partial fraction. In this question, first we will convert the fractional term $\dfrac{2x+1}{\left( x-2 \right)\left( {{x}^{2}}+4 \right)}$ into partial fractions that is $\dfrac{A}{x-2}$ and $\dfrac{Bx+C}{{{x}^{2}}+4}$. And, from here we will find the values of A and B to make the integration easier. After converting into partial fractions, we will integrate them.

Complete answer:
Let us solve this question.
In this question, it is asked to integrate $\int{\dfrac{2x+1}{\left( x-2 \right)\left( {{x}^{2}}+4 \right)}dx}$ using partial fraction.
As we can see here, we can use partial fraction here. Because the degree of the numerator is less than the degree of the denominator.
So, first let us find the partial fraction of the term $\dfrac{2x+1}{\left( x-2 \right)\left( {{x}^{2}}+4 \right)}$ by splitting up in two terms.
$\dfrac{2x+1}{\left( x-2 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{A}{x-2}+\dfrac{Bx+C}{{{x}^{2}}+4}$
Let us multiply $\left( x-2 \right)\left( {{x}^{2}}+4 \right)$ on both sides of the equation. We can write the above equation as
$\Rightarrow 2x+1=\left( x-2 \right)\left( {{x}^{2}}+4 \right)\dfrac{A}{x-2}+\left( x-2 \right)\left( {{x}^{2}}+4 \right)\dfrac{Bx+C}{{{x}^{2}}+4}$
$\Rightarrow 2x+1=\left( {{x}^{2}}+4 \right)A+\left( x-2 \right)\left( Bx+C \right)$
Now, let us distribute the factors by multiplying with the parenthesis
$\Rightarrow 2x+1=A{{x}^{2}}+4A+B{{x}^{2}}+Cx-2Bx-2C$
The above equation also can be written as
$\Rightarrow 2x+1=A{{x}^{2}}+B{{x}^{2}}+Cx-2Bx-2C+4A$
As we know that if a number or a variable is multiplied with zero, then the multiplication will be zero.
In the above equation, we will add $0\times {{x}^{2}}$ to the left hand side of the equation to make the equation balanced and find the values of A, B, and C.
$\Rightarrow 0\times {{x}^{2}}+2x+1=A{{x}^{2}}+B{{x}^{2}}+Cx-2Bx-2C+4A$
$\Rightarrow 0\times {{x}^{2}}+2x+1=\left( A+B \right){{x}^{2}}+\left( C-2B \right)x-2C+4A$
Now, let us compare the coefficients of square of x, compare the coefficient of x and also compare the constants.
A+B=0 : Equation 1
C-2B=2 : Equation 2 and
1=-2C+4A : Equation 3
From equation 1, we can say that A=-B or B=-A
Putting the value of A=-B in equation 2, we get
C+2A=2 or 2 =C+2A
Multiplying 2 in both sides of the above equation, we get that
4=2C+4A : Equation 4
Now, adding equation 3 and equation 4, we get
5=8A
$\Rightarrow A=\dfrac{5}{8}$
From here, we get that
$B=-A=-\dfrac{5}{8}$
By putting the values of B in equation 2 that is C-2B=2, we get
$C-2\times \left( -\dfrac{5}{8} \right)=2$
$\Rightarrow C+\dfrac{5}{4}=2$
$\Rightarrow C=2-\dfrac{5}{4}=\dfrac{3}{4}$
So, we can write the equation $\dfrac{2x+1}{\left( x-2 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{A}{x-2}+\dfrac{Bx+C}{{{x}^{2}}+4}$ by putting the values of A, B, and C, we get
$\dfrac{5}{8\left( x-2 \right)}+\dfrac{\dfrac{-5}{8}x+\dfrac{3}{4}}{{{x}^{2}}+4}=\dfrac{5}{8\left( x-2 \right)}+\dfrac{-5x+6}{8\left( {{x}^{2}}+4 \right)}$
Now, we integrate them
$\int{\dfrac{2x+1}{\left( x-2 \right)\left( {{x}^{2}}+4 \right)}dx}=\int{\left( \dfrac{5}{8\left( x-2 \right)}+\dfrac{-5x+6}{8\left( {{x}^{2}}+4 \right)} \right)dx}$
$\Rightarrow \int{\dfrac{2x+1}{\left( x-2 \right)\left( {{x}^{2}}+4 \right)}dx}=\int{\left( \dfrac{5}{8\left( x-2 \right)} \right)dx}+\int{\left( \dfrac{-5x}{8\left( {{x}^{2}}+4 \right)} \right)dx}+\int{\left( \dfrac{6}{8\left( {{x}^{2}}+4 \right)} \right)dx}$
Taking out the constants from the integration, we get
$\Rightarrow \int{\dfrac{2x+1}{\left( x-2 \right)\left( {{x}^{2}}+4 \right)}dx}=\dfrac{5}{8}\int{\dfrac{1}{\left( x-2 \right)}dx}+\dfrac{(-5)}{8}\int{\dfrac{x}{\left( {{x}^{2}}+4 \right)}dx}+\dfrac{6}{8}\int{\dfrac{1}{\left( {{x}^{2}}+4 \right)}dx}$
We are going to use some formulas here. The formulas are:
$\int{\dfrac{1}{x-a}dx}=\ln (x-a)$
$\int{\dfrac{x}{{{x}^{2}}+{{a}^{2}}}dx=}\dfrac{1}{a}\ln \left( {{x}^{2}}+{{a}^{2}} \right)$
$\int{\dfrac{1}{\left( {{x}^{2}}+{{a}^{2}} \right)}dx}=\dfrac{1}{a}{{\tan }^{-1}}\dfrac{x}{a}$
Using the above formula, we can write the integration as
$\Rightarrow \int{\dfrac{2x+1}{\left( x-2 \right)\left( {{x}^{2}}+4 \right)}dx}=\dfrac{5}{8}\ln (x-2)+\dfrac{(-5)}{8}\dfrac{1}{2}\ln \left( {{x}^{2}}+4 \right)+\dfrac{6}{8}\dfrac{1}{2}{{\tan }^{-1}}\dfrac{x}{2}+C$
By simplifying the above equation, we can write
$\Rightarrow \int{\dfrac{2x+1}{\left( x-2 \right)\left( {{x}^{2}}+4 \right)}dx}=\dfrac{5}{8}\ln (x-2)-\dfrac{-5}{16}\ln \left( {{x}^{2}}+4 \right)+\dfrac{3}{8}{{\tan }^{-1}}\dfrac{x}{2}+C$

Note: We should have a better knowledge in integration, mainly a partial fraction topic for solving this type of question. Some formulas should be kept remembered for solving this type of question easily. The formulas we have used in this solution are:
$\int{\dfrac{1}{x-a}dx}=\ln (x-a)$
$\int{\dfrac{x}{{{x}^{2}}+{{a}^{2}}}dx=}\dfrac{1}{a}\ln \left( {{x}^{2}}+{{a}^{2}} \right)$
$\int{\dfrac{1}{\left( {{x}^{2}}+{{a}^{2}} \right)}dx}=\dfrac{1}{a}{{\tan }^{-1}}\dfrac{x}{a}$
And don’t forget to add a constant C after the integration.