((Δbe▽)/(math)) ∫ ((arc tan x)/((1+(1/x^2 )))) dx ?


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Question Number 109891 by bemath last updated on 26/Aug/20

$\frac{Δ b e ▽}{m a t h}$ $\int \frac{arc \tan x}{(1 + \frac{1}{x^{2}})} d x ?$
	Answered by john santu last updated on 26/Aug/20
	$((JS)/(★■★)) I=∫ ((x^2 tan^(−1) (x))/(x^2 +1)) dx = ∫ (({(x^2 +1)−1}tan^(−1) (x))/(x^2 +1)) dx = ∫ (((x^2 +1)tan^(−1) (x))/(x^2 +1))dx−∫ ((tan^(−1) (x))/(x^2 +1)) dx =∫ tan^(−1) (x) dx−∫ tan^(−1) (x) d(tan^(−1) (x)) = x.tan^(−1) (x)−∫ (x/(x^2 +1)) dx−(1/2)(tan^(−1) (x))^2 =x.tan^(−1) (x)−(1/2)ln (x^2 +1)−(1/2)(tan^(−1) (x))^2 + c$
	$\frac{J S}{★ ◼ ★}$ $I = \int \frac{x^{2} \tan^{- 1} (x)}{x^{2} + 1} d x$ $= \int \frac{{(x^{2} + 1) - 1} \tan^{- 1} (x)}{x^{2} + 1} d x$ $= \int \frac{(x^{2} + 1) \tan^{- 1} (x)}{x^{2} + 1} d x - \int \frac{\tan^{- 1} (x)}{x^{2} + 1} d x$ $= \int \tan^{- 1} (x) d x - \int \tan^{- 1} (x) d (\tan^{- 1} (x))$ $= x . \tan^{- 1} (x) - \int \frac{x}{x^{2} + 1} d x - \frac{1}{2} {(\tan^{- 1} (x))}^{2}$ $= x . \tan^{- 1} (x) - \frac{1}{2} \ln (x^{2} + 1) - \frac{1}{2} {(\tan^{- 1} (x))}^{2} + c$
	Commented by bemath last updated on 26/Aug/20
	a farmer math
	Commented by Rasheed.Sindhi last updated on 26/Aug/20

	$D o y o u s e e a n y h o e s t h e r e ? :)$
	Commented by bemath last updated on 26/Aug/20

	$n o s i r . h a h a h a h a$
	Commented by Rasheed.Sindhi last updated on 26/Aug/20
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