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explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 52

Answers (1)

Answer:

                \log \left|\frac{(x-3)^{7}}{(x-2)^{5}}\right|+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

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

Explanation:

Let

\begin{aligned} &I=\int \frac{2 x+1}{(x-2)(x-3)} d x \\ &\frac{2 x+1}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3} \end{aligned}

(2 x+1)=A(x-3)+B(x-2)                                 (1)

Put x= 3

Equation (1)

\begin{aligned} &6+1=B \\ &B=7 \end{aligned}

Put x= 2

Equation (1)

\begin{aligned} &4+1=A(-1) \\ &A=-5 \\ &\frac{2 x+1}{(x-2)(x-3)}=\frac{-5}{x-2}+\frac{7}{x-3} \end{aligned}

\begin{aligned} &I=-5 \int \frac{d x}{x-2}+7 \int \frac{d x}{x-3} \\ &I=-5 \log |x-2|+7 \log |x-3|+C \\ &I=\log \left|\frac{(x-3)^{7}}{(x-2)^{5}}\right|+C \end{aligned}

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