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Question: How do you solve \[\int {\dfrac{1}{{{x^2}\left( {x - 1} \right)}}dx} \]...

How do you solve ∫1x2(x−1)dx\int {\dfrac{1}{{{x^2}\left( {x - 1} \right)}}dx}

Explanation

Solution

We are given functions in the ratio form. Since the function is not directly mentioned we will use the partial fractions method. In which we will split the given integral into parts. Then we will find the values of or real values of the splitted numerators. And on simplified fractions we will find the integral.

Complete step by step answer:
Given that the integral is,
∫1x2(x−1)dx\int {\dfrac{1}{{{x^2}\left( {x - 1} \right)}}dx}
Now we will decompose this integral by partial fractions,
1x2(x−1)=Ax+Bx2+Cx−1\dfrac{1}{{{x^2}\left( {x - 1} \right)}} = \dfrac{A}{x} + \dfrac{B}{{{x^2}}} + \dfrac{C}{{x - 1}}
Now we will find the LCM of the terms on RHS,
1x2(x−1)=Ax.(x−1)x2(x−1)+B.(x−1)x2(x−1)+C.x2x2(x−1)\dfrac{1}{{{x^2}\left( {x - 1} \right)}} = \dfrac{{Ax.\left( {x - 1} \right)}}{{{x^2}\left( {x - 1} \right)}} + \dfrac{{B.\left( {x - 1} \right)}}{{{x^2}\left( {x - 1} \right)}} + \dfrac{{C.{x^2}}}{{{x^2}\left( {x - 1} \right)}}
Since the denominators on both sides are the same. So we can easily equate the numerators,
1=Ax.(x−1)+B.(x−1)+C.x21 = Ax.\left( {x - 1} \right) + B.\left( {x - 1} \right) + C.{x^2}
This is the simplified form. Now we will find the respective values of A, B and C.
Putting the value of x=0:
−1=B- 1 = B
Putting the value of x=1:
1=C1 = C
Now we will put these values of B and C in the simplified equations above.
1=Ax.(x−1)−1(x−1)+x21 = Ax.\left( {x - 1} \right) - 1\left( {x - 1} \right) + {x^2}
Multiplying the brackets one by one,
1=Ax2−Ax−x+1+x21 = A{x^2} - Ax - x + 1 + {x^2}
Taking the terms with A as coefficient on one side,
1+x−1−x2=Ax2−Ax1 + x - 1 - {x^2} = A{x^2} - Ax
On simplifying we get,
x−x2=A(x2−x)x - {x^2} = A\left( {{x^2} - x} \right)
Taking minus sign common from the LHS terms,
−(x2−x)=A(x2−x)- \left( {{x^2} - x} \right) = A\left( {{x^2} - x} \right)
Cancelling the common terms on both sides we get,
−1=A- 1 = A
This is the value of the last variable.
Now the integral becomes,
∫1x2(x−1)dx=∫−1xdx+∫−1x2dx+∫1x−1dx\int {\dfrac{1}{{{x^2}\left( {x - 1} \right)}}dx = \int {\dfrac{{ - 1}}{x}dx} + \int {\dfrac{{ - 1}}{{{x^2}}}dx} + \int {\dfrac{1}{{x - 1}}} } dx
These are the standard integrals.
So using formulas of integration ∫−1xdx\int {\dfrac{{ - 1}}{x}dx} and ∫−1x2dx\int {\dfrac{{ - 1}}{{{x^2}}}dx}
Now taking the integration we get,
∫1x2(x−1)dx=−ln⁡∣x∣+1x+ln⁡∣x−1∣+C\int {\dfrac{1}{{{x^2}\left( {x - 1} \right)}}dx = - \ln \left| x \right| + \dfrac{1}{x} + \ln \left| {x - 1} \right| + C}
We know the standard integrals in ln form so we can write
∫1x2(x−1)dx=ln⁡∣x−1x∣+1x+C\int {\dfrac{1}{{{x^2}\left( {x - 1} \right)}}dx = } \ln \left| {\dfrac{{x - 1}}{x}} \right| + \dfrac{1}{x} + C

Note:
To solve these problems we need to know the integral formulas. Also note that this type of way to solve the integral is used when you want to convert the given integral into standard forms. Students should always cross check after the partial fraction whether they did it correctly or not.