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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 30

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 30

Answers (1)

Answer:

            \frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

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

Explanation:

Let

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

Equating the similar terms

\begin{aligned} &A+C=0 \\ &A=-C \end{aligned}                         (1)

A+B-2C=1            (2)

\begin{aligned} &A+B+2 A=1 \\ &B+3 A=1 \end{aligned}                    (3)

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

Adding equation (3) and (4)

\begin{aligned} &3 B=1 \\ &B=\frac{1}{3} \end{aligned}

Equation (3)

\begin{aligned} &\frac{1}{3}+3 A=1 \\ &3 A=1-\frac{1}{3} \\ &3 A=\frac{2}{3} \\ &A=\frac{2}{9} \\ &C=\frac{-2}{9} \end{aligned}

\frac{x}{(x-1)^{2}(x+2)}=\frac{2}{9(x-1)}+\frac{1}{3(x-1)^{2}}+\frac{-2}{9(x+2)}

\begin{aligned} &I=\frac{2}{9} \int \frac{d x}{x-1}-\frac{2}{9} \int \frac{d x}{x+2}+\frac{1}{3} \int \frac{d x}{(x-1)^{2}} \\ &I=\frac{2}{9} \log |x-1|-\frac{2}{9} \log |x+2|-\frac{1}{3(x-1)}+C \\ &I=\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C \end{aligned}

Posted by

infoexpert27

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