Question

$\frac{dy}{dx}=x^5{\tan }^{-1}\left(x^3\right)$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=x^5{\tan }^{-1}\left(x^3\right) \\ \Rightarrow dy=\left\{x^5{\tan }^{-1}\left(x^3\right)\right\}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int x^5{\tan }^{-1}\left(x^3\right)dx \\ \Rightarrow y=\int x^5{\tan }^{-1}\left(x^3\right)dx \\ \mathrm{Putting}t=x^3,\mathrm{we}\mathrm{get} \\ dt=3x^{2}dx \\ \therefore y=\frac{1}{3}\int t{\tan }^{-1}tdt \\ =\frac{1}{3}\int \underset{\mathrm{II}}{t}\times \underset{\mathrm{I}}{{\tan }^{-1}t}dx \\ =\frac{1}{3}\left[{\tan }^{-1}t\int tdt-\int \left\{\frac{d}{dt}\left({\tan }^{-1}t\right)\int tdx\right\}dt\right] \\ =\frac{1}{3}\times \frac{t^2{\tan }^{-1}t}{2}-\frac{1}{6}\int \frac{t^2}{\left(1+t^2\right)}dt \\ =\frac{t^2{\tan }^{-1}t}{6}-\frac{1}{6}\int \frac{t^2+1-1}{\left(1+t^2\right)}dt \\ =\frac{t^2{\tan }^{-1}t}{6}-\frac{1}{6}\int dt+\frac{1}{6}\int \frac{1}{1+t^2}dt \\ =\frac{t^2{\tan }^{-1}t}{6}-\frac{1}{6}t+\frac{{\tan }^{-1}t}{6}+C \\ =\frac{x^6{\tan }^{-1}{x}^3}{6}-\frac{1}{6}{x}^3+\frac{{\tan }^{-1}{x}^3}{6}+C \\ =\frac{1}{6}\left(x^6{\tan }^{-1}{x}^3-x^3+{\tan }^{-1}{x}^3\right)+C \\ \mathrm{Hence},y=\frac{1}{6}\left(x^6{\tan }^{-1}{x}^3-x^3+{\tan }^{-1}{x}^3\right)+C\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}. \end{gathered}$$