Question

What is$\int {\tan }^{-1}xdx$ ?

Answer 1

Integration of$\mathbf{\int }{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{dx}$:

Let

$$\begin{gathered} t={\tan }^{-1}x \\ \Rightarrow \tan t=x \\ \Rightarrow {\sec }^{2}tdt=dx \end{gathered}$$

Now, $\int {\tan }^{-1}xdx=\int t{\sec }^{2}tdt$

Integrate using: $\mathbf{\int }\mathbf{u}\mathbf{}\mathbf{v}\mathbf{}\mathbf{dx}\mathbf{=}\mathbf{u}\mathbf{\int }\mathbf{}\mathbf{v}\mathbf{}\mathbf{dx}\mathbf{-}\mathbf{\int }\mathbf{}\left(\frac{\mathrm{du}}{\mathrm{dx}}\int v\mathrm{dx}\right)\mathbf{}\mathbf{dx}$

$$\begin{gathered} \int {\tan }^{-1}xdx=\int t{\sec }^{2}tdt \\ =t\tan t-\int \frac{dt}{dt}\tan tdt \\ =t\tan t-\int \tan tdt\mathbf{}\left[\because \int \mathrm{tant}\mathrm{dt}=\ln \left|\mathrm{sect}\right|\right] \\ =t\tan t-\log \left|\sec t\right|+c \\ =t\tan t-\log \left|\sqrt{1+{\tan }^{2}t}\right|+c\left[\because {\sec }^{2}t=1+{\tan }^{2}t\right] \\ =x{\tan }^{-1}x-\log \left|\sqrt{1+x^2}\right|+c \end{gathered}$$

Thus,the value of integration is$\mathbf{\int }\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathit{x}\mathbf{}\mathit{d}\mathit{x}\mathbf{=}\mathit{x}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathit{x}\mathbf{-}\mathit{l}\mathit{o}\mathit{g}\left|\sqrt{1+x^2}\right|\mathbf{+}\mathit{c}$ .