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Question: Evaluate the integral \(\int{\dfrac{{{x}^{2}}-1}{{{\left( x-1 \right)}^{2}}\left( x+3 \right)}dx}.\)...

Evaluate the integral ∫x2−1(x−1)2(x+3)dx.\int{\dfrac{{{x}^{2}}-1}{{{\left( x-1 \right)}^{2}}\left( x+3 \right)}dx}..

Explanation

Solution

In this we will evaluate the integral  ∫x2−1(x−1)2(x+3)dx \text{ }\int{\dfrac{{{x}^{2}}-1}{{{\left( x-1 \right)}^{2}}\left( x+3 \right)}dx}\text{ } by using method of partial fractions. In the method of partial fraction the function which is to be integrated should be rational function or rational algebraic function. We first represent rational function in partial form then by integral we can evaluate. Now we will give one of the general conversion of rational form into partial form which as follows:
px+q(x−a)(x−b)=Ax−a+Bx−b\dfrac{px+q}{\left( x-a \right)\left( x-b \right)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}

Complete step by step answer:
Let I=∫x2−1(x−1)2(x+3)dxI=\int{\dfrac{{{x}^{2}}-1}{{{\left( x-1 \right)}^{2}}\left( x+3 \right)}dx}
Since a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)
x2−12=(x−1)(x+1){{x}^{2}}-{{1}^{2}}=\left( x-1 \right)\left( x+1 \right)
⇒I=∫(x−1)(x+1)(x−1)2(x+3)dx\Rightarrow I=\int{\dfrac{\left( x-1 \right)\left( x+1 \right)}{{{\left( x-1 \right)}^{2}}\left( x+3 \right)}dx}
By cancelling (x-1) by both numerator and denominator, we get
⇒I=∫(x+1)(x−1)(x+3)dx ....(1)\Rightarrow I=\int{\dfrac{\left( x+1 \right)}{\left( x-1 \right)\left( x+3 \right)}dx}\text{ }....\text{(1)}
Now consider, (x+1)(x−1)(x+3)=Ax−1+Bx+3 ...(2)\dfrac{\left( x+1 \right)}{\left( x-1 \right)\left( x+3 \right)}=\dfrac{A}{x-1}+\dfrac{B}{x+3}\text{ }...\text{(2)}
By cross multiplication, we get
(x+1)=A(x+3)+B(x−1) ...(3)\left( x+1 \right)=A\left( x+3 \right)+B\left( x-1 \right)\text{ }...\text{(3)}
Put x=-3 in equation (3), we get
(−3+1)=A(−3+3)+B(−3−1)\left( -3+1 \right)=A\left( -3+3 \right)+B\left( -3-1 \right)
−2=A(0)+B(−4) -2=A\left( 0 \right)+B\left( -4 \right)\text{ }
B(−4)= (−2)B\left( -4 \right)\text{= }\left( -2 \right)
Dividing both sides by -4, we get
B=12B=\dfrac{1}{2}.
Put x=1 in equation (3), we get
(1+1)=A(1+3)+B(1−1) \left( 1+1 \right)=A\left( 1+3 \right)+B\left( 1-1 \right)\text{ }
2=A(4)+B(0) 2=A\left( 4 \right)+B\left( 0 \right)\text{ }
A(4)=2 A\left( 4 \right)\text{=2 }
Dividing both sides by 4, we get
A=12 A\text{=}\dfrac{1}{2}\text{ }
By using values of A and B in equation (2), we get
(x+1)(x−1)(x+3)=12x−1+12x+3 \dfrac{\left( x+1 \right)}{\left( x-1 \right)\left( x+3 \right)}=\dfrac{\dfrac{1}{2}}{x-1}+\dfrac{\dfrac{1}{2}}{x+3}\text{ }
Taking 12\dfrac{1}{2} common form right hand side, we get
(x+1)(x−1)(x+3)=12(1x−1+1x+3) ...(4)\dfrac{\left( x+1 \right)}{\left( x-1 \right)\left( x+3 \right)}\text{=}\dfrac{1}{2}\left( \dfrac{1}{x-1}+\dfrac{1}{x+3} \right)\text{ }...\text{(4)}
By integrating equation (4) with respect to x,
∫(x+1)(x−1)(x+3)dx=∫12(1x−1+1x+3) dx\int{\dfrac{\left( x+1 \right)}{\left( x-1 \right)\left( x+3 \right)}dx}=\int{\dfrac{1}{2}\left( \dfrac{1}{x-1}+\dfrac{1}{x+3} \right)\text{ }dx}
∫(x+1)(x−1)(x+3)dx=12∫(1x−1+1x+3) dx\int{\dfrac{\left( x+1 \right)}{\left( x-1 \right)\left( x+3 \right)}dx}=\dfrac{1}{2}\int{\left( \dfrac{1}{x-1}+\dfrac{1}{x+3} \right)\text{ }dx}
∫(x+1)(x−1)(x+3)dx=12(∫1x−1 dx+∫1x+3 dx) ...(5)\int{\dfrac{\left( x+1 \right)}{\left( x-1 \right)\left( x+3 \right)}dx}=\dfrac{1}{2}\left( \int{\dfrac{1}{x-1}\text{ }dx}+\int{\dfrac{1}{x+3}\text{ }dx} \right)\text{ }...(5)
Since ∫1x+a dx=log⁡(x+a)+c \int{\dfrac{1}{x+a}\text{ }dx=\log (x+a)}+c\text{ }
⇒∫1x−1 dx=log⁡(x−1)+c ...(6) and\Rightarrow \int{\dfrac{1}{x-1}\text{ }dx}=\log (x-1)+c\text{ }...\text{(6) and}
∫1x+3 dx=log⁡(x+3)+c ...(7)\int{\dfrac{1}{x+3}\text{ }dx}=\log (x+3)+c\text{ }...\text{(7)}
By using equation (6) and equation (7) in equation (5),
∫(x+1)(x−1)(x+3)dx=12(log⁡(x−1)+log⁡(x+3))+c\int{\dfrac{\left( x+1 \right)}{\left( x-1 \right)\left( x+3 \right)}dx}=\dfrac{1}{2}\left( \log (x-1)+\log (x+3) \right)+c
Since log⁡a+log⁡b=log(ab)\log a+ \log b=log(ab)
Hence,
 ∫(x+1)(x−1)(x+3)dx=12log((x−1)(x+3))+c \text{ }\int{\dfrac{\left( x+1 \right)}{\left( x-1 \right)\left( x+3 \right)}dx}=\dfrac{1}{2}\text{log}\left( (x-1)(x+3) \right)\text{+c }

Note:
In this problem, one should know all the formula of integration, the rational function and if f(x) and g(x)f(x)\text{ and }g(x) are two polynomial then f(x) g(x),\dfrac{f(x)}{\text{ }g(x)}, g(x)≠0g(x)\ne 0 is called rational function.