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Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.20 Question 4 maths Textbook Solution.

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Answer: 10 \log |x-3|-5 \log |x-2|+c

Given: \int \frac{x^{2}+1}{x^{2}-5 x+6} d x

Hint: Using partial fraction and \int \frac{1}{x} d x


                \begin{aligned} &\text { Let }\\ &I=\int \frac{x^{2}+1}{x^{2}-5 x+6} d x \end{aligned}

                \frac{x^{2}+1}{x^{2}-5 x+6}=\frac{x^{2}+1}{(x-3)(x-2)} \quad\left[\begin{array}{l} x^{2}-5 x+6=x^{2}-3 x-2 x+6 \\ =x(x-3)-2(x-3) \\ =(x-3)(x-2) \end{array}\right]

               \frac{x^{2}+1}{x^{2}-5 x+6}=\frac{A}{(x-3)}+\frac{B}{(x-2)}

             Multiplying by  \left ( x-3 \right ) and \left ( x-2\right )


            Putting x = 2

           4+1=A(0)+B(-1) \Rightarrow B=-5

            Putting x = 3

            9+1=A(3-2)+B(0) \Rightarrow A=10

            \therefore \int \frac{x^{2}+1}{x^{2}-5 x+6} d x=10 \int \frac{1}{x-3} d x-5 \int \frac{1}{x-2} d x

                                                =10 \log |x-3|-5 \log |x-2|+c

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