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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 6

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 6

Answers (1)

Answer:

            \frac{1}{2} \log |x-1|-4 \log |x-2|+\frac{9}{2} \log |x-3|+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

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

Explanation:

\begin{aligned} &I=\int \frac{x^{2}}{(x-1)(x-2)(x-3)} d x \\ &\frac{x^{2}}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \quad\left[\frac{p x^{2}+q x+c}{(x-a)(x-b)(x-c)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\right] \end{aligned}        

\begin{aligned} &\frac{x^{2}}{(x-1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)} \\ &x^{2}=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \end{aligned}

\begin{aligned} &\text { At } x=2 \\ &4=0+B(1)(-1)+0 \\ &4=-B \\ &B=-4 \end{aligned}

\begin{aligned} &\text { At } x=3 \\ &9=0+0+C(2)(1) \\ &9=2 C \\ &C=\frac{9}{2} \end{aligned}

\begin{aligned} &\text { At } x=1 \\ &1=A(-1)(-2)+0+0 \\ &2 A=1 \\ &A=\frac{1}{2} \end{aligned}

Now

\begin{aligned} &\frac{x^{2}}{(x-1)(x-2)(x-3)}=\frac{1}{2(x-1)}+\frac{(-4)}{x-2}+\frac{9}{2(x-3)} \\ &I=\int\left[\frac{1}{2(x-1)}-\frac{4}{x-2}+\frac{9}{2(x-3)}\right] d x \\ &I=\frac{1}{2} \int \frac{1}{x-1} d x-4 \int \frac{1}{x-2} d x+\frac{9}{2} \int \frac{1}{x-3} d x \\ &I=\frac{1}{2} \log |x-1|-4 \log |x-2|+\frac{9}{2}|x-3|+C \end{aligned}

Posted by

infoexpert27

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