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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.30 Question 31

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.30 Question 31

Answers (1)

Answer:

            \frac{1}{4} \log |x-1|+\frac{3}{4} \log |x+1|+\frac{1}{2(x+1)}+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

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

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{2}}{(x-1)(x+1)^{2}} d x \\ &\frac{x^{2}}{(x-1)(x+1)^{2}}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x+1)^{2}} \end{aligned}

\begin{aligned} &x^{2}=A(x+1)(x-1)+B(x+1)^{2}+C(x-1) \\ &x^{2}=x^{2}(A+B)+x(2 B+C)+(-A+B-C) \end{aligned}

Equating similar terms

A+B=1                           (1)

2B+C=0                       (2)

\begin{aligned} &C=-2 B \\ &-A+B-C=0 \end{aligned}                (3)

\begin{aligned} &-A+B+2 B=0 \\ &-A+3 B=0 \end{aligned}                      (4)

Adding equation (1) and (4)

\begin{gathered} 4 B=1 \\ B=\frac{1}{4} \end{gathered}

Equation (2)

\begin{aligned} &C=-2 \times \frac{1}{4} \\ &C=\frac{-1}{2} \end{aligned}

Equation (1)

\begin{aligned} &A=1-\frac{1}{4} \\ &A=\frac{3}{4} \end{aligned}

\begin{aligned} &\frac{x^{2}}{(x-1)(x+1)^{2}}=\frac{3}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2(x+1)^{2}} \\ &I=\frac{3}{4} \int \frac{1}{x+1} d x+\frac{1}{4} \int \frac{1}{x-1} d x-\frac{1}{2} \int \frac{1}{(x+1)^{2}} d x \\ &I=\frac{3}{4} \log |x+1|+\frac{1}{4} \log |x-1|+\frac{1}{2(x+1)}+C \end{aligned}

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