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explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 56 maths

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Answer: \int \frac{x \tan ^{-1} x^{2}}{1+x^{4}} d x

Hint: Use substitution method to solve this integral.

Given:   \frac{1}{4}\left(\tan ^{-1} x^{2}\right)^{2}+c


        I=\int \frac{x \tan ^{-1} x^{2}}{1+x^{4}} d x

        \begin{aligned} &\text { Put } \tan ^{-1} x^{2}=t \Rightarrow \frac{1}{1+\left(x^{2}\right)^{2}} 2 x\; d x=d t \\ &\Rightarrow \frac{2 x}{1+x^{4}}\; d x=d t \Rightarrow d x=\frac{1+x^{4}}{2 x} d t \\ &\text { Then, } I=\int \frac{x t}{1+x^{4}} \cdot \frac{1+x^{4}}{2 x} d t=\frac{1}{2} \int t\; d t \end{aligned}

                        =\frac{1}{2} \frac{t^{1+1}}{1+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

                        \begin{aligned} &=\frac{1}{2} \cdot \frac{t^{2}}{2}+c=\frac{1}{4} t^{2}+c \\ &=\frac{1}{4}\left(\tan ^{-1} x^{2}\right)^{2}+c \quad\left[\because t=\tan ^{-1} x^{2}\right] \end{aligned}


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