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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.20 Question 7 Maths Textbook Solution.

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

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

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

Explanation: Let

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

                        =1-\frac{(4 x+1)}{x^{2}+2 x+2}

                        =1-2\left(\frac{2 x+\frac{1}{2}}{x^{2}+2 x+2}\right)

                        =1-2\left(\frac{2 x+2-2+\frac{1}{2}}{x^{2}+2 x+2}\right)

                        =1-2\left(\frac{2 x+2-2+\frac{1}{2}}{x^{2}+2 x+2}\right)

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

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

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

                                   =x-2 \log \left|x^{2}+2 x+2\right|+3 \tan ^{-1}(x+1)+c



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