Question-200048

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

Commented by 0670322918 last updated on 13/Nov/23

\int \frac{{t a n}^{- 1} (x)}{\int {t a n}^{- 1} (x) d x} d x =

f (x) = \int {t a n}^{- 1} (x) d x = {x t a n}^{- 1} (x) - \frac{1}{2} l n (x^{2} + 1) + c

f^{'} (x) = {t a n}^{- 1} (x)

\int \frac{f^{'} (x)}{f (x)} d x = l n ∣ f (x) ∣ + c

\int \frac{{t a n}^{- 1} (x)}{\int {t a n}^{- 1} (x) d x} d x = l n ∣ {x t a n}^{- 1} (x) - \frac{1}{2} l n (x^{2} + 1) + c ∣ + c_{1}

Commented by Calculusboy last updated on 13/Nov/23

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