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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.20 Question 10 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.20 Question 10 maths Textbook Solution.

Answers (1)

Answer: x-3 \log \left|x^{2}+6 x+12\right|+2 \sqrt{3} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c

Given: \int \frac{x^{2}}{x^{2}+6 x+12} d x

Hint: Using \int \frac{1}{x} d x \text { and } \int \frac{1}{1+x^{2}} d x

Explanation: Let

                        I=\int \frac{x^{2}}{x^{2}+6 x+12} d x

                        \begin{aligned} &=\int\left(1-\left(\frac{6 x+12}{x^{2}+6 x+12}\right)\right) d x \\ &=\int 1 d x-6 \int \frac{x+2}{x^{2}+6 x+12} d x \\ &=\int 1 d x-\frac{6}{2} \int \frac{2 x+4+2-2}{x^{2}+6 x+12} d x \end{aligned}

                        \begin{aligned} &=\int 1 d x-3 \int \frac{2 x+6}{x^{2}+6 x+12} d x+6 \int \frac{1}{x^{2}+6 x+(3)^{2}-(3)^{2}+12} d x \\ &=\int 1 d x-3 \int \frac{2 x+6}{x^{2}+6 x+12}+6 \int \frac{1}{(x+3)^{2}+(\sqrt{3})^{2}} d x \end{aligned}

                        \begin{aligned} &=x-3 \log \left|x^{2}+6 x+12\right|+6 \times \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c \\ &=x-3 \log \left|x^{2}+6 x+12\right|+2 \sqrt{3} \tan ^{-1}\left(\frac{x+3}{\sqrt{3}}\right)+c \end{aligned}

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