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Question Number 200048 by Calculusboy last updated on 12/Nov/23

Commented by 0670322918 last updated on 13/Nov/23

$$\int \frac{{tan}^{-1}\left(x\right)}{\int {tan}^{-1}\left(x\right)dx}dx=$$$$f\left(x\right)=\int {tan}^{-1}\left(x\right)dx={xtan}^{-1}\left(x\right)-\frac{1}{2}ln(x^2+1)+c$$$$f^{\prime }\left(x\right)={tan}^{-1}\left(x\right)$$$$\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=ln\mid f\left(x\right)\mid +c$$$$\int \frac{{tan}^{-1}\left(x\right)}{\int {tan}^{-1}\left(x\right)dx}dx=ln\mid {xtan}^{-1}\left(x\right)-\frac{1}{2}ln(x^2+1)+c\mid +c_1$$$$$$

Commented by Calculusboy last updated on 13/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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