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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 122 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 122 Maths Textbook Solution.

Answers (1)

Answer: \frac{1}{8} \log \left|\frac{x-1}{x+1}\right|-\left(\frac{3 x-4}{4(x-1)^{2}}\right)+C

Given:\int \frac{x^{2}}{(x-1)^{3}(x+1)} d x

Hint: you must know the steps to integrate by partial function

Explanation: let I=\int \frac{x^{2}}{(x-1)^{3}(x+1)} d x

\frac{x^{2}}{(x-1)^{3}(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)^{3}}+\frac{D}{(x+1)}

Multiplying by (x-1)^{3}(x+1)

x^{2}=A(x-1)^{3}(x+1)+B(x-1)(x+1)+C(x+1)+D(x-1)^{3}

Putting x=1

1=A(0)+B(0)+C(2)+D(0) \Rightarrow c=\frac{1}{2}

Putting x=-1

1=A(0)+B(0)+C(2)+D(-8) \Rightarrow c=\frac{-1}{8}

Putting x=0

\begin{aligned} &1=A(1)+B(-1)+C(1)+D(-1)\\ &A-B+C-D=0\\ &A-B+\frac{1}{2}-\left(-\frac{1}{8}\right)=0 \Rightarrow A-B+\frac{1}{2}+\frac{1}{8}=0\\ &A-B+\frac{5}{8}=0 \Rightarrow A-B=-\frac{5}{8} \end{aligned}

Putting x=2

\begin{aligned} &4=A(1)(3)+B(1)(3)+C(3)+D(1) \\ &4=3 A+3 B+3\left(\frac{1}{2}\right)+\left(-\frac{1}{8}\right) \\ &\Rightarrow 3 A+3 B=4-\frac{3}{2}+\frac{1}{8}=>3(A+B)=\frac{32-12+1}{8}=\frac{21}{8} \\ &A+B=\frac{3}{8} \end{aligned}

Adding (i) and (ii)

\begin{aligned} &A+B=\frac{-5}{8}+\frac{7}{8}=\frac{2}{8} \\ &2 A-\frac{2}{8} \Rightarrow A=\frac{1}{8} \end{aligned}

Put in (1)

\begin{aligned} &\frac{1}{8}-B=\frac{5}{8} \Rightarrow B=\frac{1}{8}+\frac{5}{8}=\frac{6}{8}=>B=\frac{3}{4} \\ &\frac{x^{2}}{(x-1)^{3}}=\frac{1}{8(x-1)}+\frac{3}{4(x-1)^{2}}=\frac{1}{2(x-1)^{3}}=\frac{1}{8(x+1)} \\ &\frac{1}{8} \int \frac{d x}{x-1}+\frac{3}{4} \int \frac{(x-1)^{-2+1}}{-2+1}+\frac{1}{2} \frac{(x-1)^{-3+1}}{-3+1}-\frac{1}{8} \log |x+1|+C \\ &=\frac{1}{8} \log \left|\frac{x-1}{x+1}\right|-\frac{3 x-4}{4(x+1)^{2}}+C \end{aligned}

 

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