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Question: How do you integrate \(\int{\dfrac{1}{{{x}^{2}}\left( 2x-1 \right)}}dx\) using partial fractions?...

How do you integrate ∫1x2(2x−1)dx\int{\dfrac{1}{{{x}^{2}}\left( 2x-1 \right)}}dx using partial fractions?

Explanation

Solution

In the given question, we have been asked to integrate the given expression using partial fraction. Firstly use the partial fraction method to express the integrand in two separate integrands whose denominators are the individual factors of the given integrand. And then integrate them separately.

Complete step by step solution:
In order to integrate ∫1x2(2x−1)dx\int{\dfrac{1}{{{x}^{2}}\left( 2x-1 \right)}}dx using partial fractions, we will first separate or express the integrand as sum of two integrands as follows;
1x2(2x−1)=Ax2+Bx2+C(2x−1)\dfrac{1}{{{x}^{2}}\left( 2x-1 \right)}=\dfrac{A}{{{x}^{2}}}+\dfrac{B}{{{x}^{2}}}+\dfrac{C}{\left( 2x-1 \right)}
Now, we will find the value of A, B and C by comparing integrands as follows
⇒1=A(2x−1)+Bx(2x−1)+C(x2)\Rightarrow 1=A\left( 2x-1 \right)+Bx\left( 2x-1 \right)+C\left( {{x}^{2}} \right)
Equating similar terms (constants and coefficients of the variable) we will get,
let x=0,⇒ 1=−A ⇒A=−1let\ x=0,\Rightarrow \ 1=-A\ \Rightarrow A=-1
let x=12,⇒ 1=C4 ⇒C=4let\ x=\dfrac{1}{2},\Rightarrow \ 1=\dfrac{C}{4}\ \Rightarrow C=4
0=2B+C ⇒B=−C2 ⇒B=−42=−20=2B+C\ \Rightarrow B=-\dfrac{C}{2}\ \Rightarrow B=-\dfrac{4}{2}=-2
Substituting the value of A, B and C.
Therefore,
Now we have the respective values of A, B and C, so putting them in the above considered integrands and then integrating them, we will get
1x2(2x−1)=−1x2−2x2+4(2x−1)\dfrac{1}{{{x}^{2}}\left( 2x-1 \right)}=-\dfrac{1}{{{x}^{2}}}-\dfrac{2}{{{x}^{2}}}+\dfrac{4}{\left( 2x-1 \right)}

Now, putting integration to each term of both the sides,
Thus, integrating the resultant expression, we obtain
∫1x2(2x−1)dx=−∫1x2dx−∫2x2dx+∫4(2x−1)dx\int{\dfrac{1}{{{x}^{2}}\left( 2x-1 \right)}}dx=-\int{\dfrac{1}{{{x}^{2}}}dx}-\int{\dfrac{2}{{{x}^{2}}}}dx+\int{\dfrac{4}{\left( 2x-1 \right)}}dx
Now, we know that the integration of 1(x±a)\dfrac{1}{(x\pm a)} equalsln⁡∣x±a∣+c\ln \left| x\pm a \right|+c, so integrating the above functions using this formula, we will get
∴∫1x2(2x−1)dx=1x−2ln⁡∣x∣+2ln⁡(∣2x−1∣)+C\therefore \int{\dfrac{1}{{{x}^{2}}\left( 2x-1 \right)}}dx=\dfrac{1}{x}-2\ln \left| x \right|+2\ln \left( \left| 2x-1 \right| \right)+C
Hence, it is the required integration.

Note: When doing indefinite integration, always write +C part after the integration. This +C part indicates the constant part remains after integration and can be understood when you explore it graphically. The finite integration constant gets cancelled out, so we only write it in indefinite integration. We should remember the property or the formulas of integration, this would make it easier to solve the question. You should always remember all the methods for integration so that we can easily choose which method is suitable for solving the particular type of question.
Partial fraction of the rational fraction form p(x)+q(x±a)(x±b)\dfrac{p(x)+q}{(x\pm a)(x\pm b)} is given as A(x±a)+B(x±b)\dfrac{A}{(x\pm a)}+\dfrac{B}{(x\pm b)}
Find the value of A and B by comparing it with original integrand.