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Q.(i). $\int \frac{x \cos ec\left({\tan }^{-1}{x}^2\right)}{1+x^4}dx$.

Dear Student,

$$\begin{gathered} Let I = \int \frac{x \cos ec \left({\tan }^{-1}{x}^2\right)}{1 + x^4} dx \\ Let {\tan }^{-1}{x}^2 = t \\ \Rightarrow \frac{1}{1 + {\left(x^2\right)}^2} \times 2x = \frac{dt}{dx} \\ \Rightarrow \frac{x}{1 + x^4}dx = \frac{dt}{2} \\ \therefore I = \int \frac{x \cos ec \left({\tan }^{-1}{x}^2\right)}{1 + x^4} dx \\ = \int \frac{\cos ec t}{2}dt \\ = \frac{1}{2}\int \cos ec t dt \\ = \frac{1}{2}\log \left|\cos ec t - cot t\right| + C \\ = \boxed{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{log}\left|\mathbf{cos}\mathbf{e}\mathbf{c}\left({\mathbf{tan}}^{\mathbf{-}\mathbf{1}}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{c}\mathbf{o}\mathbf{t}\left({\mathbf{tan}}^{\mathbf{-}\mathbf{1}}{\mathbf{x}}^{\mathbf{2}}\right)\right|\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{C}} \end{gathered}$$

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