∫ ((xtan^(−1) (x))/(√(1+x^2 ))) dx ?


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Question Number 104104 by bemath last updated on 19/Jul/20

$\int \frac{x \tan^{- 1} (x)}{\sqrt{1 + x^{2}}} d x ?$
	Answered by OlafThorendsen last updated on 19/Jul/20

	$By parts :$ $\sqrt{1 + x^{2}} \arctan x - \int \sqrt{1 + x^{2}} \frac{d x}{1 + x^{2}}$ $\sqrt{1 + x^{2}} \arctan x - \int \frac{d x}{\sqrt{1 + x^{2}}}$ $\sqrt{1 + x^{2}} \arctan x - argsh x + C$
	Answered by Dwaipayan Shikari last updated on 19/Jul/20
	$∫((θtanθ)/(secθ)) sec^2 θdθ {x=tanθ ⇒1=sec^2 θ(dθ/dx) ∫θsecθtanθ θ∫secθtanθ−∫∫secθtanθ θsecθ−log(secθ+tanθ)+C tan^(−1) x sec(tan^(−1) x)−log(sec(tan^(−1) x)+x)+C Another way tan^(−1) x.(1/2)∫((2x)/(√(x^2 +1)))−∫(1/(x^2 +1)).(1/2)∫((2x)/(√(x^2 +1))) tan^(−1) x(√(x^2 +1))−∫(dx/(√(x^2 +1)))=tan^(−1) x (√(x^2 +1))−log(x+(√(x^2 +1)))+C$
	$\int \frac{θ t a n θ}{s e c θ} {s e c}^{2} θ d θ {x = t a n θ \Rightarrow 1 = {s e c}^{2} θ \frac{d θ}{d x}$ $\int θ s e c θ t a n θ$ $θ \int s e c θ t a n θ - \int \int s e c θ t a n θ$ $θ s e c θ - l o g (s e c θ + t a n θ) + C$ ${t a n}^{- 1} x s e c ({t a n}^{- 1} x) - l o g (s e c ({t a n}^{- 1} x) + x) + C$ $A n o t h e r w a y$ ${t a n}^{- 1} x . \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} + 1}} - \int \frac{1}{x^{2} + 1} . \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} + 1}}$ ${t a n}^{- 1} x \sqrt{x^{2} + 1} - \int \frac{d x}{\sqrt{x^{2} + 1}} = {t a n}^{- 1} x \sqrt{x^{2} + 1} - l o g (x + \sqrt{x^{2} + 1}) + C$
	Answered by bobhans last updated on 19/Jul/20

	$\int \tan^{- 1} (x) d (\sqrt{1 + x^{2}}) =$ $\sqrt{1 + x^{2}} \tan^{- 1} (x) - \int \frac{\sqrt{1 + x^{2}}}{1 + x^{2}} d x =$ $\sqrt{1 + x^{2}} \tan^{- 1} (x) - \int \frac{d x}{\sqrt{1 + x^{2}}}$ $l e t I_{1} = \int \frac{d x}{\sqrt{1 + x^{2}}} [x = \tan p]$ $I_{1} = \int \frac{\sec^{2} p}{\sec p} d p = \int \sec p d p$ $I_{1} = \ln ∣ \sec p + \tan p ∣ + c$ $∴ I = \sqrt{1 + x^{2}} \tan^{- 1} (x) - \ln ∣ \sqrt{1 + x^{2}} + x ∣ + c$
	Answered by mathmax by abdo last updated on 19/Jul/20

	$I = \int \frac{xarctanx}{\sqrt{1 + x^{2}}} dx by parts u^{^{'}} = \frac{x}{\sqrt{1 + x^{2}}} and v = arctanx \Rightarrow$ $I = \sqrt{1 + x^{2}} arctanx - \int \sqrt{1 + x^{2}} \times \frac{dx}{1 + x^{2}} = arctanx \sqrt{1 + x^{2}} - \int \frac{dx}{\sqrt{1 + x^{2}}}$ $= \arctan (x) \sqrt{1 + x^{2}} - \ln (x + \sqrt{1 + x^{2}}) + C$
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